


















































































































































































RAILROAD CURVES 


AND 


EARTHWORK 

*■ - T \ 

' * 


BY 


C. FRANK ALLEN, S.B. 

11 

MEMBER AMERICAN SOCIETY OF CIVIL ENGINEERS 
PROFESSOR OF RAILROAD ENGINEERING IN THE MASSACHUSETTS 

INSTITUTE OF TECHNOLOGY 


SEVENTH THOUSAND 
FOURTH EDITION, REVISED 

i * > 


NEW YORK 

SPON & CHAMBERLAIN, 123 LIBERTY STREET. 

LONDON 

E. & F. N. SPON, Ltd., 125 STRAND 

1907 



f 






I 


LIBRARY of CONGRESS 
l wo Copies Received 

SEP 9 I 90 f 

Cooyricht Biitry 

A<<*f /<6rfo 7 

CLAsM XXC./N0, 
COPY B. 





Copyright, 1SS9, 1894, 1903, and 1907, 
By C. F. ALLEN. 



Norfajooh -Press 

J. S. Cushing Co. — Berwick & Smith Co. 
Norwood, Mass., U.S.A. 



PREFACE. 



This book was prepared for the use of the students in the 
author’s classes. It has been used in lithographed sheets for a 
number of years in very nearly the present form, and has given 
satisfaction sufficient to suggest putting it in print. An effort 
has been made to have the demonstrations simple and direct, 
and special care has been given to the arrangement and the 
typography, in order to secure clearness and conciseness of 
mathematical statement. Much of the material in the earlier 
part of the book is necessarily similar to that found in one or 
more of several excellent field books, although the methods of 
demonstration are in many cases new. This will be found true 
especially in Compound Curves, for which simple treatment 
has been found quite possible. New material will be found in 
the chapters on Turnouts and on “Y” Tracks and Crossings. 
The Spiral Easement Curve is treated originally. The chapters 
on Earthwork are essentially new ; they include Staking Out; 
Computation, directly and with Tables and Diagrams; also 
Haul, treated ordinarily and by Mass Diagram. Most of the 
material relating to Earthwork is not elsewhere readily available 
for students’ use. 

The book has been written especially to meet the needs of 
students in engineering colleges, but it is probable that it will 
be found useful by many engineers in practice. The size of 
page allows it to be used as a pocket book in the field. It is 
difficult to avoid typographical and clerical errors ; the author 
will consider it a favor if he is notified of any errors found to 
exist. 

C. FRANK ALLEN. 

Boston, 

September, 1899. 

ill 


5 


PREFACE TO THIRD EDITION. 

In this edition the chapter on the Spiral Easement Curve 
has been entirely re-written and enlarged; the treatment is 
largely new as to detail. The author desires, however, to ac¬ 
knowledge his indebtedness to the paper on Transition Curves, 
by W. B. Lee, in the Transactions of the American Society of 
Civil Engineers, December, 1901. Two pages have been added 
to the chapter on Special Problems in Earthwork. The refer¬ 
ences to Tables have been changed so as to apply to the newly 
published Tables by the author of this book. 

C. FRANK ALLEN. 

July, 1903. 


PREFACE TO FOURTH EDITION. 

The chapter on Turnouts has been almost entirely re-written, 
and now applies especially to the Split Switch. The treatment 
is largely original. The changes in this chapter have carried 
with them changes in the treatment of “ Y ” tracks. It is be¬ 
lieved that these chapters are now up to date. 

The chapter on Spirals has been re-arran$ed and a number 
of illustrated examples added, but the subject-matter is little 
changed except in the arrangement. 

The chapter on Reversed Curves has been re-arranged ; some 
of the demonstrations have been simplified, others made more 
general. 

Throughout the book a page has been re-written here and 
there, where experience in teaching has shown the desirability 
of making some point a trifle clearer. 

While the number of pages is increased by only fourteen, 
nearly sixty pages are new or have been fully re-set, while 
minor corrections have been made in the plates of many others. 

C. FRANK ALLEN. 


July, 1907. 


iv 



CONTENTS 


CHAPTER I. 

Reconnoissance. 

SECTION . PAGE 

1. Operations in location. 1 

2. Reconnoissance. 1 

3. Nature of examination. 1 

4. Features of topography. 2 

5. Purposes of reconnoissance. 3 

6. Elevations, how taken. 3 

7. Pocket instruments used. 4 

8. Importance of reconnoissance. 5 

CHAPTER II. 

Preliminary Survey. 

9. Nature of preliminary. 6 

10. Grades. 6 

11. Importance of low grades. 7 

12. Pusher grades. 7 

13-14. Purposes of preliminary... 8 

15. Nature. 9 

16. Methods. 9 

17. Backing up. 10 

18. Notes. H 

19. Organization of party. 11 

20. Locating engineer. 11 

21. Transitman ; also form of notes. 12 

22. Head chainman. 13 

23. Stakeman. 13 

24. Rear chainman. 14 

25. Back flag. 14 

26. Axeman. 14 

27. Leveler; also form of notes. 14 


V 




























Vlll 


Contents. 


SECTION PAGE 

110 . Given, long chord, angles, and R s ; required Ii, Is, 7, Ri • 58 

111. Given, long chord, angles, and Ri; required 7/, Is, I, Rs . 58 

112. Substitute for simple, a compound curve to end in par¬ 

allel tangent. 58 

113. Example. 59 

114. Change P.C.C. so as to join parallel tangent . 00 

115. Substitute for simple, a symmetrical curve with flattened 

ends. 61 

110. Substitute curve with flattened ends to pass through 

middle point.v. 62 

117. Substitute simple curve for curves with connecting tangent 63 

CHAPTER VI. 

Reversed Curves. 

118. Use of reversed curves. 04 

119-122. Between parallel tangents, common radius.04-05 

123-124. Between parallel tangents, unequal radii.65-00 

125-126. Find 7 ls 7 2 , 7 2 when 7, 7\, R lf R . 2 are given.66-67 

127-128. Find common radius to connect tangents not parallel.67-08 

CHAPTER VII. 

Parabolic Curves. 

129. Use of parabolic curves.7*.. 69 

130. Properties of the parabola. 69 

131. Lay out parabola by offsets from tangent. 70 

132. Field-work. 71 

133. Parabola by middle ordinates. 72 

134. Vertical curves, where used . 72 

135. Method for vertical curve 200 feet long. 73 

136. General method. 74 

137. Example. 7 <; 

138. To find proper length of vertical curve. 76 

CHAPTER VIII. 

Turnouts. 

139. Definitions. 77 

140. Find frog angle from number of frog. 78 























Contents. 


IX 


BF.CTION PAGE 

141-142. Description of split switch . 79 

142. Find radius of turnout for split switch. 80 

143. Radius of turnout for split switch and straight frog. 81 

144. Turnout beyond frog. 82 

145. Methods of connecting parallel tracks by turnouts.82-83 

146-148. Formulas for simple cases.83-84 

149. Reference curve. 85 


150. Comparison of radii, reference curve and split switch .. 
151-152. Additional cases of parallel tangents with turnouts. 
153-154. Formulas for a series of parallel tracks. 

155. Examples. 

156. 

157. 

158. 

159. 


. 86 
.88-89 
. 90 

. 91 

Crotch frog for split switches . 92 

Modified reference curve defined. 92 

Radius for modified reference curve inside. 93 

Approximate formula for above . 94 

160-161. Modified reference curve outside.95-96 

161. Example of case § 161. 96 

162. Bending process described. 97 

163. Find radius of turnout curve from frog to parallel curved 

track outside; also approximate method.98-99 

164. Example of precise method. 99 

165. Example of approximate method. 10° 

166. Radius of turnout curve from frog to parallel curved track 

inside. *90 

167. Special case of turnout outside. 101 

168. Cross-over between parallel curved tracks. 101 

169. Stub switch described. 102 

170-171. Stub switch formulas, including crotch frog. 103 


CHAPTER IX. 


“ y ” Tracks and Crossings. 


172. Definition. 

173. Main track tangent, “Y” track curved, and turnout 

curved.. .. 

174. Main track tangent, “ Y ” curved, turnout curved with 

tangent . 

175-176. Main track tangent and curve “ Y ” curved, turnout 

curved . 

177. Crossing of tangent and curve. 

178. Crossing of two curves. 


104 

104 

105 

106 

107 

108 
































X 


Contents. 


CHAPTER x. 

Cubic Spiral Easement Curve. 

SECTION PAGE 

179. Necessity for; also elevation of outer rail. 109 

180. Equations for cubic parabola and cubic spiral.110-111 

181. Discussion of character of easement curves. Ill 

182. Deflection, angles, spiral angles, and properties of spirals 112 

183. Values of y in terms of l and R c . 113 

184. Values of p and q in terms of x, ?/, Rc, and s c . 113 

185. Tangent distances, with example. 114-115 

186. Laying out when D c and p are given, with example .. .116-117 
187-188. Laying out when D c and l c are given ; also field-work 118 

189. Laying out cubic spiral by offsets, with example. 119 

190-191. Deflection angles from intermediate points.120-121 

192. Spirals for compound curves... 122 

193. Example; also field-work . 123 

194-196. Substitution of curve and spiral for simple curve. .124-126 


CHAPTER XI. 

Setting Stakes for Earthwork. 


197. Data. 127 

198. What stakes and how marked. 127 

199. Method of finding rod reading for grade. 128 

200. Example. 129 

201. Cut or fill at center. 129 

202. Side stake for level section ........\ . 130 

203-206. Side stakes when surface is not level.130-132 

207. Slope-board or level-board. 132 

208-210. Keeping the notes. 133 

211-212. Form of note-book.13L-135 

213. Cross-sections; where taken . 136 

214-215. Passing from cut to fill.136-137 

216. Opening in embankment. 137 

217. General level notes. 137 

218-221. Level, three-level, five-level, irregular sections . 138 


CHAPTER XII. 

Methods of computing Earthwork. 


222. Principal methods used. 130 

223. Averaging end areas. 130 





























Contents. 


xi 


SECTION PACE 

224. Kinds of cross-sections specified. 140 

225. Level cross-section. 140 

226. Three-level section. 141 

227. Three-level section; second method. 142 

228. Five-level section . 148 

229. Irregular section. 113 

230. Planimeter . 144 

231. Comment on end area formula. 144 

232. Prismoidal formula. 144 

233. Prismoidal formula for prisms, wedges, pyramids . 145 

234. Natui’e of regular section of earthwork. 144 

235-237. Prismoidal formula applied where upper surface is 

warped. 146-148 

238. "Wide application of prismoidal formula. 148 

239-240. Prismoidal correction .149-150 

241. Where applicable; also special case. 151 

242. Correction for pyramid. 152 

243. Correction for five-level sections. 152 

244-245. Correction for irregular sections .152-153 

246. Value of prismoidal correction. 153 

247. Method of middle areas. 154 

248. Method of equivalent level sections. 154 

249. Method of mean proportionals. 154 

250. Henck’s method. 154 

251. Formula. 155 

252-254. Example.156-157 

255-256. Comment on Henck’s and end area methods.157-158 

257-263. Example comparing the various methods. 158 

CHAPTER XIII. 

Special Problems in Earthwork. 

264. Correction for curvature . 159 

265. Correction where chords are less than 100 feet. 161 

266. Correction of irregular sections. 161 

267. Opening in embankment. 162 

268. Borrow-pits. 114 

269. Truncated triangular prism. 164 

270. Truncated rectangular prism. 165 

271. Assembled prisms. 167 

272. Additional heights. 168 






































Xll 


Contents. 


CHAPTER XIV. 

Earthwork Tables. 

section page 

273. Formula for use in tables . 170 

274. Arrangement of table. 171 

275. Explanation of table. 171 

276-277. Example of use, including prismoidal correction table 172 

278. Prismoidal correction applied for section less than 100 feet 173 

279. Tables, where published. 173 

280. Tables of triangular prisms. 173 

281. Where published. 173 

282. Arrangement of tables of triangular prisms. 174 

283. Example of use. 175 

284-285. Application to irregular sections. 176 

CHAPTER XV. 

Earthwork Diagrams. 

286-287. Method of diagrams. 177 

288. Forms of equations available for straight lines. 178 

289. Method of use of diagrams.. 178 

290-291. Computations and table for diagram of prismoidal 

correction . 179 

292. Diagram for prismoidal correction and explanation of 

construction.^. 180 

293. Explanation and example of use. 182 

294. Table of triangular prisms. 182 

295-298. Computations and table for diagram of three-level 

sections. 183 

299-300. Checks upon computations.\ 187 

301. Explanation of diagram; also curve of level section. 187 

302. Use of diagram for three-level sections. 188 

303. Comment on rapidity by use of diagrams. 189 

304. Special use to find prismoidal correction for irregular 

sections. 189 

CHAPTER XVI. 

Haul. 

305. Definition and measure of haul. 190 

306-307. Length of haul, how found. 190 


























Contents. 


X1U 


SECTION PAGE 

308. Formula for center of gravity of a section. 101 

309-310. Formula deduced. 102 

311. Formula modified for use with tables or diagrams. 104 

312. For section less than 100 feet. 104 

313. For series of sections. 195 

» 

CHAPTER XVII. 

Mass Diagram. 

314. Definition. 196 

315. Table and method of computation . 196 

316. Mass diagram and its properties. 198 

317. Graphical measure of haul explained. 199 

318. Application to mass diagram. 201 

319. Further properties. 201 

320. Mass diagram; showing also borrow and waste. 203 

321. Profitable length of haul. 203 

322-323. Example of use of diagram. 205 

324. Effect of shrinkage on mass diagram . 206 

325. Discussion of overhaul . 206 

326. Treatment of overhaul by mass diagram. 207 


Tables and Diagrams 


208-220 


























. 















• ^ 

. 

. 

d >, #jh ,. ■ 

' 
















Qj 'Sl III 






























RAILROAD CURVES AND EARTHWORK. 



CHAPTER I. 


1. The operations of “locating ’ y a railroad, as commonly 
practiced in this country, are three in number : — 

I. Reconnoissance. 

II. Preliminary Survey. 

III. Location Survey. 


I. RECONNOISSANCE. 

2. The Reconnoissance is a rapid survey, or rather a critical 
examination of country, without the use of the ordinary instru¬ 
ments of surveying. Certain instruments, however, are used, 
the Aneroid Barometer, for instance. It is very commonly the 
case that the termini of the railroad are fixed, and often inter¬ 
mediate points also. It is desirable that no unnecessary re¬ 
strictions as to intermediate points should be imposed on the 
engineer to prevent his selecting what he finds to be the best 
line, and for this reason it is advisable that the reconnoissance 
should, where possible, precede the drawing of the charter. 

3. The first step in reconnoissance should be to procure the 
best available maps of the country ; a study of these will gen¬ 
erally furnish to the engineer a guide as to the routes or section 
of country that should be examined. If maps of the United 
States Geological Survey are at hand, with contour lines and 
other topography carefully shown, the reconnoissance can be 
largely determined upon these maps. Lines clearly imprac¬ 
ticable will be thrown out, the maximum grade closely deter¬ 
mined, and the field examinations reduced to a minimum No 

1 




2 


Railroad Carves and Earthwork. 


route should be. accepted finally from any such map, but a 
careful field examination should be made over the routes indi¬ 
cated on the contour maps. The examination, in general, 
should cover the general section of country, rather than be 
confined to a single line between the termini. A straight line 
and a straight grade from one terminus to the other is desirable, 
but this is seldom possible, and is in general far from possible. 
If a single line only is examined, and this is found to be nearly 
straight throughout, and with satisfactory grades, it may be 
thought unnecessary to carry the examination further..*ft will* 
frequently, however, be found advantageous to deviate con¬ 
siderably from a straight line in order to secure satisfactory 
grades. In many cases it will be necessary to wind about more 
or less through the country in order to secure the best line. 
Where a high hill or a mountain lies directly between the 
points, it may be expected that a line around the hill, and 
somewhat remote from a direct line, will prove more favorable 
than any other. Unless a reasonably direct line is found, the 
examination, to be satisfactory, should embrace all the section of 
intervening country, and all feasible lines should be examined. 

4. There are two features of topography that are likely to 
prove of especial interest in reconnoissance, ridge lines and 
valley lines. 

A ridge line along the whole of its course is higher than the 
ground immediately adjacent to it on each side. That is, the 
ground slopes downward from it to both sides. It is also called 
a watershed line. 

A valley line , to the contrary, is lower than the ground im¬ 
mediately adjacent to it on each side. The ground slopes 

up waul from it to both sides. Valley lines may be called water¬ 
course lines. 

A pass is a place on a ridge line lower than any neighboring 
points on the same ridge. Very important points to be deter¬ 
mined in reconnoissance are the passes where the ridge lines 
are to be crossed; also the points where the valleys are to be 
crossed; and careful attention should be given to these points. 
In crossing a valley through which a large stream flows, it may 
be of great importance to find a good bridge crossing. In some 
cases where there are serious difficulties in crossing a ridge, a 
tunnel may be necessary. Where such structures, either 


Reconnoismnce. 


3 


bridges or tunnels, are to be built, favorable points for their 
construction should be selected and the rest of the line be com¬ 
pelled to conform. In many parts of the United States at the 
present time, the necessity for avoiding grade crossings causes 
the crossings of roads and streets to become governing points 
of as great importance as ridges and valleys. 

5. There are several purposes of reconnoissance: first, to 
find whether there is any satisfactory line between the proposed 
termini; if so, second, to establish which is the most feasible; 
third, to determine approximately the maximum grade neces¬ 
sary to be used ; fourth. report upon the character or 
geological formation of the country, and the probable cost of 
construction depending somewhat upon that ; fifth, to make 
note of the existing resources of the country, its manufactures, 
mines, agricultural or natural products, and the capabilities for 
improvement and development of the country resulting from 
the introduction of the railroad. The report upon reconnois¬ 
sance should include information upon all these points. It is 
for the determination of the third point mentioned, the rate of 
maximum grade, that the barometer is used. Observing the 
elevations of governing points, and knowing the distances be¬ 
tween those points, it is possible to form a good judgment as to 
what rate <>f maximum grade to assume. 

6. The Elevations are usually taken by the Aneroid Barome¬ 
ter. Tables for converting barometer readings into elevations 
above sea-level are readily available and in convenient form for 
field use. (See Searles’ or Henck’s Field Books.) 

Distances may be determined with sufficient accuracy in 
many cases from the map, where a good one exists. Where 
this method is impossible or seems undesirable, the distance 
may be determined in one of several different ways. When 
the trip is made by wagon, it is customary to use an Odometer , 
an instrument which measures and records the number of 
revolutions of the wheel to which it is attached, and thus the 
distance traveled by the wagon. There are different forms of 
odometer. In its most common form, it depends upon a hang¬ 
ing weight or pendulum, which is supposed to hold its position, 
hanging vertical, while the wheel turns. The instrument is 
attached to the wheel between the spokes and as near to the 
hub as practicable. At low speeds it registers accurately ; as the 


4 


Railroad Curves and Earthwork. 


speed is increased, a point is reached where the centrifugal force 
neutralizes or overcomes the force of gravity upon the pendu¬ 
lum, and the instrument fails to register accurately, or perhaps 
at high speeds to register at all. If this form of odometer is 
used, a clear understanding should be had of the conditions 
under which it fails to correctly register. A theoretical discus¬ 
sion might closely establish the point at which the centrifugal 
force will balance the force of gravity. The wheel striking 
against stones in a rough road will create disturbances in the 
action of the pendulum, so that the odometer will fail to register 
accurately at speeds less than that determined upon the above 
assumption. 

Another form of odometer is manufactured which is con¬ 
nected both with the wheel and the axle, and so measures 
positively the relative motion between the wheel and axle, 
and this ought to be reliable for registering accurately. Many 
engineers prefer to count the revolutions of the wheel them¬ 
selves, tying a rag to the wheel to make a conspicuous mark 
for counting. 

When the trip is made on foot, pacing will give satisfactory 
results. An instrument called the Pedometer registers the 
results of pacing. As ordinarily constructed, the graduations 
read to quarter miles, and it is possible to estimate to one- 
tenth that distance. Peddmeters are also made which register 
paces. In principle, the pedometer depends upon the fact that, 
with each step, a certain shock or jar is produced as the heel 
strikes the ground, and each shock causes the instrument to 
register. Those registering miles are adjustable to the length 
of pace of the wearer. 

If the trip is made on horseback, it is found possible to get 
good results with a steady-gaited horse, by first determining his 
rate of travel and figuring distance by the time consumed in 
traveling. Excellent results are said to have been secured in 
this way. 

7. It is customary for engineers not to use a compass in 
reconnoissance, although this is sometimes done in order to 
trace the line traversed upon the map, and with greater accu¬ 
racy. A pocket level will be found useful. The skillful use of 
pocket instruments will almost certainly be found of great valua 
to the engineer of reconnoissance. 


Reconnoissance. 


5 


It may, in cases, occur that no maps of any value are in 
existence or procurable. It may be necessary, in such a case, 
o make a rapid instrumental survey, the measurements bein^ 

ta en either by pacing, chain, or stadia measurements. This 
is, however, unusual. 

8 . The preliminary survey is based upon the results of the 
reconnoissance, and the location upon the results of the pre¬ 
liminary survey. The reconnoissance thus forms the founda¬ 
tion upon which the location is made. Any failure to find a 
suitable line and the best line constitutes a defect which no 
amount of faithfulness in the later work will rectify. The 
most serious errors of location are liable to be due to imper¬ 
fect reconnoissance; an inefficient engineer of reconnoissance 
should be avoided at all hazards. In the case of a new railroad, 
it would, in general, be proper that the Chief Engineer should 
in person conduct this survey. In the case of the extension of 
existing lines, this might be impracticable or inadvisable, but 
an assistant of known responsibility, ability, and experience 
should in this case be selected to attend to the work. 


CHAPTER II. 


II. PRELIMINARY SURVEY. 

9. The Preliminary Survey is based upon the results of the 
reconnoissance. It is a survey made with the ordinary instru¬ 
ments of surveying. Its purpose is to fix and mark upon the 
ground a first trial line approximating as closely to the proper 
final line as the difficulty of the country and the experience of 
the engineer will allow ; further than this, to collect data such 
that this survey shall serve as a basis upon which the final 
Location may intelligently be made. In order to approximate 
closely in the trial line, it is essential that the maximum grade 
should be determined or estimated as correctly as possible, and 
the line fixed with due regard thereto. 

It will be of value to devote some attention here to an ex¬ 
planation about Grades ami “Maximum Grades.” 

10. Grades. — The ideal line in railroad location is a straight 
and level line. This is seldom, if ever, realized. When the two 
termini are at different elevations, a line straight and of uni¬ 
form grade becomes the ideal. It is commonly impossible to 
secure a line of uniform grade between termini. In operating 
a railroad, an engine division w r ill be about 100 miles, some¬ 
times less, often more. In locating any 100 miles of railroad, 
it is almost certain that a uniform grade cannot be maintained. 
More commonly there will be a succession of hills, part of the 
line up grade, part down grade. Sometimes there will be a 
continuous up grade, but not at a uniform rate. With a uni¬ 
form grade, a locomotive engine will be constantly exerting its 
maximum pull or doing its maximum work in hauling the 
longest train it is capable of hauling; there will be no power 
wasted in hauling a light train over low or level grades upon 
which a heavier train could be hauled. Where the grades are 
not uniform, but are rising or falling, or rising irregularly, it 
will be found that the topography on some particular 5 or 10 

0 


Preliminary Survey. 


7 


miles is of such a character that the grade here must be steeper 
than is really necessary anywhere else on the line ; or there 
may be two or three stretches of grade where about the same 
rate of grade is necessary, steeper than elsewhere required. 
The steep grade thus found necessary at some special point or 
points on the line of railroad is called the “ Maximum Grade” 
or “Ruling Grade” or “Limiting Grade,” it being the grade 
that limits the weight of train that an engine can haul over the 
whole division. It should then be the effort to make the rate 
of maximum grade as low as possible, because the lower the 
rate of the maximum grade, the heavier the train a given loco¬ 
motive can haul, and because it costs not very much more to 
haul a heavy train than a light one. The maximum grade 
determined by the reconnoissance should be used as the basis 
for the preliminary survey. How will this affect the line ? 
Whenever a hill is encountered, if the maximum grade be 
steep, it may be possible to carry the line straight, and over 
the hill; if the maximum grade be low, it. may be necessary to 
deflect the line and carry it around the hill. When the maxi¬ 
mum grade has been once properly determined, if any saving 
can be accomplished by using it rather than a grade less steep, 
the maximum grade should be used. It is possible that the 
train loads will not be uniform throughout the division. It 
will be advantageous to spend a small sum of money to keep 
any grade lower than the maximum, in view of the possibility < 
that at this particular point the train load will be heavier than 
elsewhere on the division. Any saving made will in general 
be of one or more of three kinds : — 

a. Amount or “ quantity ” of excavation or embankment; 

b. Distance; 

c. Curvature. 

11 . In some cases, a satisfactory grade, a low grade for a 
maximum, can be maintained throughout a division of 100 
miles in length, with the exception of 2 or 3 miles at one point 
only. So great is the value of a low maximum grade that all 
kinds of expedients will be sought for, to pass the difficulty 
without increasing the rate of maximum grade, which we know 
will apply to the whole division. 

12 . Sometimes by increasing the length of line, we are able 
to reach a given elevation with a lower rate of grade. Some- 


8 


Railroad Curves and Earthwork. 


times heavy and expensive cuts and fills may serve the pur¬ 
pose. Sometimes all such devices fail, and there still remains 
necessary an increase of grade at this one point, but at this 
point only. In such case it is now customary to adopt the 
higher rate of grade for these 2 or 3 miles and operate them by 
using an extra or additional engine. In this case, the “ruling 
grade ” for the division of 100 miles is properly the “maximum 
grade” prevailing over the division generally, the higher grade 
for a few miles only being known as an “ Auxiliary Grade” or 
more commonly a “ Pusher Grade.” The train which is hauled 
over the engine division is helped over the auxiliary or pusher 
grade by the use of an additional engine called a “Pusher.” 
Where the use of a short “ Pusher Grade ” will allow the use 
of a low “ maximum grade,” there is evident economy in its 
use. The critical discussion of the importance or value of 
saving distance, curvature, rise and fall, and maximum grade, 
is not within the scope of this book, and the reader is referred 
to Wellington’s “ Economic Theory of Railway Location.” 

13 . The Preliminary Survey follows the general line marked 
out by the reconnoissance, but this rapid examination of coun¬ 
try may not have fully determined which of two or more lines 
is the best, the advantages may be so nearly balanced. In this 
case two or more preliminary surveys must be made for com¬ 
parison. When the reconnoissance has fully determined the 
general route, certain details are still left for the preliminary 
survey to determine. It may be necessary to run two lines, 
one on each side of a small stream, and possibly a line crossing 
it several times. The reconnoissance would often fail to settle 
minor points like this. It is desirable that the preliminary 
survey should closely approximate to the final line, but it is 
not important that it should fully coincide anywhere. 

An important purpose of the “preliminary” is to provide a 
map which shall show enough of the topography of the country, 
so that the Location proper may be projected upon this map. 
Working from the line of survey as a base line, measurements 
should be taken sufficient to show streams and various natural 
objects as well as the contours of the surface. 

14 . The Preliminary Survey serves several purposes : — 

First. To fix accurately the maximum grade for use in 

Location. 


Preliminary Survey. 


9 


Second. To determine which of several lines is best. 

Third. To provide a map as a basis upon which the Location 
can properly be made. 

Fourth. To make a close estimate of the cost of the work. 

Fifth. To secure, in certain cases, legal rights by filing plans. 

15 . It should be understood that the preliminary survey 
is, in general, simply a means to an end, and rapidity and 
economy are desirable. It is an instrumental survey. Meas¬ 
urements of distance are taken usually with the chain, although 
a tape is sometimes used. Angles are taken generally with a 
transit; some advocate the use of a compass. The line is 
ordinarily run as a broken line with angles, but is occasionally 
run with curves connecting the straight stretches, generally for 
the reason that a map of such a line is available for filing, and 
certain legal rights result from such a filing. With a compass, 
no backsight need be taken, and, in passing small obstacles, a 
compass will save time on this account. A transit line can be 
carried past an obstacle readily by a zigzag line. Common 
practice among engineers favors the use of the transit rather 
than the compass. Stakes are set, at every “ Station,” 100 feet 
apart, and the stakes are marked on the face, the first 0, the 
next I , then 2, and so to the end of the line. A stake set 1025 
feet from the beginning would be marked 10 + 25. 

Levels are taken on the ground at the side of the stakes, and 
as much oftener as there is any change in the inclination of the 
ground. All the surface heights are platted on a profile, and 
the grade line adjusted. 

16 . The line should be run from a governing point towards 
country allowing a choice of location, that is from a pass or 
from an important bridge crossing, towards country offering no 
great difficulties. There is an advantage in running from a 
summit downhill, subject, however, to the above considera¬ 
tions. In running from a summit down at a prescribed rate 
of grade, an experienced engineer will carry the line so that, at 
the end of a day’s work, the levels will show the line to be 
about where it ought to be. For this purpose, the levels must 
be worked up and the profile platted to date at the close of 
each day. Any slight change of line found necessary can then 
be made early the next morning. A method sometimes adopted 
in working down from a summit is for the locating engineer to 


10 


Railroad Carves and Eartltworlc. 


plat his grade line on the profile, daily in advance, and then 
during the day, plat a point on his profile whenever he can 
conveniently get one from his leveler, and thus find whether 
his line is too high or too low. 

17 . Occasionally the result of two or three days’ work will 
yield a line extremely unsatisfactory, enough so that the work 
of these two or three days will be abandoned. The party 
“backs up” and takes a fresh start from some convenient 
point. In such case the custom is not to tear out several 
pages of note-book, but instead to simply draw a line across 
the page and mark the page “Abandoned.” At some future 
time the abandoned notes may convey useful information to 
the effect that this line was attempted and found unavailable. 
In general, all notes worth taking are worth saving. 

Sometimes after a line has been run through a section of 
country, there is later found a shorter or better line. 


CM 



if) 

In the figure used for illustration, the first line, “ A ” Line, 
is represented by AEBCD, upon which the stations are marked 
continuously from A to D, 850 stations. The new line, “ B ” 
Line, starts from E, Sta. 102 -f 60, and the stationing is held 
continuous from 0 to where it connects with the “A” Line at 
C. The point C is Sta. 312 -f 27 of the "A” Line, and is also 
Sta. 307 + 13 of the “B” Line. It is not customary to restake 
the line from C to D in accordance with “ B ” Line stationing. 
Instead of this, a note is made in the note-books as follows : — 

Sta. 312 + 27 “ A ” Line = 307 + 13 “ B ” Line. 

Some engineers make the note in the following form : — 

Sta. 307 to 313 = 86 ft. 

The first form is preferable, being more direct and less liable 
to cause confusion. 


Preliminary Survey. 


11 


13. All notes should be kept clearly and nicely in a. note¬ 
book— never on small i>ieces of paper. The date and the 
names of members of the party should be entered each day in 
the upper left-hand corner of the page. An office copy should 
be made as soon as opportunity offers, both for safety and con¬ 
venience. The original notes should alicays he preserved; they 
would be admissible as evidence in a court of law where a copy 
would be rejected. When two or more separate or alternate 
lines are run, they may be designated 

Line “ A,” Line “ B,” Line “ C,” 
or " A ” Line, “ B ” Line, “ C ” Line. 


^ Transit Tarty. 


19. The Organization of Party may be as follows : — 

1. Locating Engineer. 

2. Transitman. 

;{. Head Chainman. 

4. Stakeman. 

5. Hear Chainman. 

G. Hack Flag. 

7. Axemen (one or more). 

8. Leveler. / 

9. Rodman (sometimes two). 1 

30. Topographer. / 

11. Assistant. J 

12. Cook. 

13. Teamster. 


Level Party. 
Topographical Party 


20. The Locating Engineer is the chief of party, and is 
responsible for the business management of the camp and 
party, as well as for the conduct of the survey. He deter¬ 
mines where the line shall run, keeping ahead of the transit, 
and establishing points as foresights or turning-points for the 
transitman. In open country, the extra axeman can assist by 
holding the flag at turning-points, and thus allowing the locat¬ 
ing engineer to push on and pick out other points in advance. 
The locating engineer keeps a special note-book or memorandum 
book ; in it he notes on the ground the quality of material, rock, 
earth, or whatever it may be ; takes notes to determine the 
lengths and positions of bridges, culverts, and other structures ; 
shows the localities of timber, building stones, borrow pits, and 




12 


Railroad Curves and Earthwork. 

other materials valuable for the execution of the work ; m fact, 
makes notes of all matters not properly attended to by the 
transit, leveling, or topography party. The rapid and faithful 
prosecution of the work depend upon the locating engineer, 
and the party ought to derive inspiration from the energy and 
vigor of their chief, who should be the leader in the work. In 
open and easy country, the locating engineer may instill life 
into the party by himself taking the place of the head chain- 
man occasionally. In country of some difficulty, his time will 
be far better employed in prospecting for the best line. 

21. The Transitman does the transit work, ranges in the 
line from the instrument, measures the angles, and keeps the 
notes of the transit survey. The following is a good form for 
the left-hand page of the note-book : — 


Station 

Point 

Deflection 

Observed 

Bearing 

Calculated 

Bearing 

7 

6 

0 + 24 

33°02'R 

N 3° 30' E 

N 3° 38' E 

5 



N 29° 30' W 

N 29° 24'W 

4 

© 

12° 09' L 



3 





2 

1 



N 17° 15'W 

N 17° 15'W 

0 

0 





Notes of topography and remarks are entered on the right- 
hand page, which, for convenience, is divided into small squares 
by blue lines, with a red line running up and down through the 
middle. 

The stations run from bottom to top of page. The bearing is 
taken at each setting and recorded just above the corresponding 
point in the note-book, or opposite a part of the line, rather 
than opposite the point. Ordinarily, the transitman takes the 
bearings of all fences and roads crossed by the line, finds the 
stations from the rear chainman, and records them in their 
proper place and direction on the right-hand page of the note¬ 
book. Section lines of the United States Land Surveys should be 












Preliminary Survey. 


13 


observed in the same way. The transitman is next in authority 
to the locating engineer, and directs the work when the latter 
is not immediately present. The transitman, while moving 
from point to point, setting up, and ranging line, limits the 
speed of the entire party, and should waste no time. 

22. The Head Chainman carries a “flag” and the forward 
end of the chain, which should be held level and Arm with one 
hand, while the flag is moved into line with the other. He 
should always put himself nearly in line before receiving a 
signal from the transitman ; plumbing may be done with the 
flag. When the point is found, the stakeman will set the stake. 
When a suitable place for a turning-point is reached, a signal 
should be given the transitman to that effect. A nail should be 
set in top of the stake at all turning-points. A proper under¬ 
standing should be had with the transitman as to signals. 

Signals from the Transitman. 

A horizontal movement of the hand indicates that the rod 
should be moved as directed. 

A swinging movement of the hand, “Plumb the rod as 
indicated.” 

A movement of both hands, or waving the handkerchief 
freely above the head, means “All right.” 

At long distances, a handkerchief can be seen to advantage; 
when snow is on the ground, something black is better. 

Signals from the Head Chainman. 

Setting the bottom of flag on the ground and waving the top, 
means “ Give the line.” 

Raising the flag above the head and holding it horizontal 
with both hands : “ Give line for a turning-point.” 

The “all right” signal is the same as from the transitman. 

In all measurements less than 100 feet (or a full chain), the 
head chainman holds the end of the chain, leaving the reading 
of the measurement to the rear chainman. 

The head chainman regulates the speed of the party during 
the time that the instrument is in place, and should keep alive 
all the time. The rear chainman will keep up as a matter of 
necessity. 

23. The Stakeman carries, marks, and drives the stakes at 
the points indicated by the head chainman. The stakes should 


14 


Railroad Carves and Earthwoi'k. 


be driven with the flat side towards the instrument, and marked 
on the front with the number of the station. Intermediate 
stakes should be marked with the number of the last station 
-f the additional distance in feet and tenths, as 10 + 67.4. The 
stationing is not interrupted and taken up anew at each turning- 
point, but is continuous from beginning to end of the survey. 
At each turning-point a plug should be driven nearly flush with 
the ground, and a witness stake driven, in an inclined position, 
at a distance of about 15 inches from the plug, and at the side 
towards which the advance line deflects, and marked W and 
under it the station of the plug. 

24. The Rear Chainman holds the rear end of the chain 
over the stake last set, but does not hold against the stake to 
loosen it. He calls “ Chain ” each time when the new stake is 
reached, being careful not to overstep the distance. He should 
stand beside the line (not on it) when measuring, and take pains 
not to obstruct the view of the transitman. He checks, and is 
responsible for the correct numbering of stakes, and for all 
distances less than 100 feet, as the head chainman always holds 
the end of the chain. The stations where the line crosses fences, 
roads, and streams should be set down in a small note-book, and 
reported to the trarrsitman at the earliest convenient opportu¬ 
nity. The rear chainman is responsible for the chain. 

25. The Back Flag holds the flag as a backsight at the 
point last occupied by the transit. The only signals necessary 
for him to understand from the transitman are “plumb the 
flag” and “all right.” The flag should always be in position, 
and the transitman should not be delayed an instant. The 
back flag should be ready to come up the instant he receives 
the “all right” signal from the transitman. The duties are 
simple, but frequently are not well performed. 

26. The Axeman cuts and clears through forest or brush. 
A good axeman should be able to keep the line well, so as to 
cut nothing unnecessary. In open country, he prepares the 
stakes ready for the stakeman or assists the locating engineer 
as fore flag. 

27. The Leveler handles the level and generally keeps the 
notes, which may have the following form for the left-hand 
page. The right-hand page is for remarks and descriptions of 
turning-points and bench-marks. It is desirable that turning- 


Preliminary Survey. 15 


Station 

+ s 

H 1 

- s 

Elevation 

B.M. 

4.67 

104.67 


100.00 

0 



5.7 

99.0 

1 



6.9 

97.8 

2 



3.4 

101.3 

T.P. 

9.26 

112.81 

1.12 

103.55 

3 



8.5 

104.3 


points should, where possible, be described, and that all bench¬ 
marks should be used as turning-points. Headings on turning- 
points should be recorded to hundredths or to thousandths of 
a foot, dependent upon the judgment of the Chief Engineer. 
Surface readings should be made to the nearest tenth, and ele¬ 
vations set down to nearest tenth only. A self-reading rod has 
advantages over a target rod for short sights. A target rod is 
possibly better for long sights and for turning-points. The 
“Philadelphia Rod” is both a target rod and a self-reading 
rod, and is thus well adapted for railroad use. Bench-marks 
should be taken at distances of from 1000 to 1500 feet, depending 
upon the country. All bench-marks, as soon as calculated, 
should be entered together On a special page near the end of 
the book. The leveler should test his level frequently to see 
that it is in adjustment. T he leveler and rodman should 
together bring the notes to date every evening and plat the 
profile to correspond. 

The profile of the preliminary line should show : — 

a. Surface line (in black). 

b. Grade line (in red). 

c. Grade elevations at each change in grade (in red). 

d. Rate of grade, per 100 (in red); rise +, fall -. 

e. Station and deflection at each angle in the line (in black). 

/. Notes of roads, ditches, streams, bridges, etc. (in black). 

28. The Rodman carries the rod and holds it vertical upon 
the ground at each station and at such intermediate points as 
mark any important change of slope of the ground. The sur¬ 
face of streams and ponds should be taken when met, and at 
frequent intervals where possible, if they continue near the line, 












16 


Railroad Curves and Earthwork. 

5 

Levels should also be taken of high-water marks wherever 
traces of these are visible. The rodman carries a small note¬ 
book in which he enters the rod readings at all turning-points. 
In country which is open, but not level, the transit party is 
liable to outrun the level party. In such cases greater speed 
will be secured by the use of two rodmen. 

29. The Topographer is, or should be, one of the most val¬ 
uable members of the party. In times past it has not always 
been found necessary to have a topographer, or if employed, 
his duty has been to sketch in the general features necessary to 
make an attractive map, and represent hills and buildings suffi¬ 
ciently well with reference to the line to show, in a general 
way, the reason for the location adopted. Sometimes the chief 
of the party has for this purpose taken the topography. At 
present the best practice favors the taking of accurate data by 
the topography party. 

The topographer (with one or two assistants) should take the 
station and bearing (or angle) of every fence or street line 
crossed by the survey (unless taken by the transit party); also 
take measurements and bearings for platting all fences and 
buildings near enough to influence the position of the Location; 
also sketch, as well as may be, fences, buildings, and other 
topographical features of interest .which are too remote to re¬ 
quire exact measurement; and Anally, establish the position of 
contour lines, streams, and ponds, within limits such that the 
Location may be well and fully determined in the contoured 
map. 

The work of taking contours is accomplished by the use of 
hand level and tape (pacing may, in many cases, be sufficient). 
The elevation at the center line at each station is found from 
the leveler. Contours are generally taken at vertical intervals 
of 5 feet. The contour line, say to the right of the station, is 
found by reading upon a light hand-marked rod the difference 
in elevation between the center line at the stake, and the re¬ 
quired contour. A cloth tape, held by the assistant topog¬ 
rapher, will serve the purpose of a rod. The next contour to 
the right can be readily found, owing to the fact that the topog¬ 
rapher’s eye is nearly 5 feet above the ground, making leveling 
easy. An allowance should be made for the difference between 
5 feet and the height of the eye. Sections both right and left 


Preliminary. Survey. 


17 


should be taken as often as necessary, the distances to each con¬ 
tour measured, and the lines between the points thus determined 
sketched in on the ground. Books of convenient size are made 
and divided into small cross-sections to facilitate sketching. 
Cross-section blocks or pads will be considered equally good by 
some engineers. The distance to which contour lines should be 
taken depends on the character of the country. The object 
should be to take contours as far from the line as is necessary 
in order to furnish contours requisite for determining the posi¬ 
tion of the located line. 

Instead of a hand level, some engineers use a clinometer and 
take and record side slopes. 

Topography can be taken rapidly and well by stadia survey 
or by plane table. This is seldom done, as most engineers are 
not sufficiently familiar with their use. 

30. Some engineers advocate making a general topographi¬ 
cal survey of the route by stadia, instead of the survey above 
described. In this case no staking out by “ stations ” would be 
done. All points occupied by the transit should be marked by 
plugs, which can be used to aid in marking the Location on 
the ground after it is determined on the contour map. This 
method has been used a number of times, and is claimed to give 
economical and satisfactory results; it is probable that it will 
have constantly increasing use in the future, and will prove 
the best method in a large share of cases. 


III. LOCATION SURVEY. 


31. The Location Survey is the final fitting of the line to 
the ground. In Location, curves are used to connect the straight 
lilies or “tangents,” and the alignment is laid out complete, 
ready for construction. 

The party is much the same as in the preliminary, and the 
duties substantially the same. More work devolves upon the 
transitman on account of the curves, and more skill is useful 
in the head cliainman in putting himself in position on curves. 
He can readily range himself on tangent. The form of notes 
will be shown later. The profile is the same, except that it 
shows, for alignment notes, the P.C. and P.T. of curves, and also 
the degree and central angle, and whether to the right or left. 

It is well to connect frequently location stakes with prelimi¬ 
nary stakes, when convenient, as a check on the work. 

In making the location survey, two distinct methods are in 
use among engineers : — 

32. First Method of Location. —Use preliminary survey and 
preliminary profile as guides in reading the country, and locate 
the line upon the ground. Experience in such work will enable 
an engineer to get very satisfactory results in this way, in 
nearly all cases. The best engineers, in locating in this way, 
as a rule lay the tangents first, and connect the curves after¬ 
wards. It will appear later how this is done. 

33. Second Method. —Use preliminary line, preliminary pro- 
fil ', and especially the contour lines on the preliminary map; 
make a paper location, and run this in on the ground. Some 
go so far as to give their locating engineer a complete set of 
notes to run by. This is going too far. Whether it is best to 
go farther than to fix, on the map, the location of tangents, 
and specify the degree of curve, is a question, A conservative 
method Is to do no more than this, and in some cases, leave 

18 


Location Survey . 


19 


he degree of curve even an open question. The second method 
is gaining in favor, hut the first method is, even now, much 
used. It is well accepted, among engineers, that no reversed 
curve should he used ; 200 feet of tangent, at least, should inter¬ 
vene. Neither should any curve be very short, say less than 
300 feet in length. 


34. A most difficult matter is the laying of a long tangent, 
so that it shall be straight. Lack of perfect adjustment and 
construction of instrument will cause a “swing” in the tan¬ 
gent. The best way is to run for a distant foresight. Another 
way is to have the transit as well adjusted as possible, and even 
then change ends every time in reversing, so that errors shall 
not accumulate. It will be noticed that the preliminary is run 
in without curves because more economical in time ; sometimes 
curves are run however, either because the line can be lun 
closer to its proper position, or sometimes in order to allow of 
filing plans with the United States or separate States. 

35. In Location, a single tangent often takes the place of a 
broken line in the preliminary, and it becomes important to 
determine the direction of the tangent with reference to some 
part of the broken line. This is readily done by finding the 
coordinates of any given point with reference to that part of 
the broken line assumed temporarily as a meridian, the 
course of each line is calculated, and the coordinates of any 
point thus found. It simplifies the calculation to use some 
part of the preliminary as an assumed meridian, rather than to 
use the actual bearings of the lines. The coordinates of two 
points on the proposed tangent allow the direction of the 
tangent to be determined with reference to any part of the 
preliminary. When the angles are small, an approximation 
sufficiently close will be secured, by assuming in all cases that 
the cosine of the angle is 1.000000 and that the smes are directly 
proportional to the angles themselves. In addition to this take 
the distances at the nearest even foot, and the calculation 

becomes much simplified. . . , . 

36 The located line, or “ Location,” as it is often called, is 

staked out ordinarily by center stakes which mark a succession 
of straight lines, connected by curves to winch the straight ines 
are tangent. The straight lines are by general usage called 

“ Tangents.” 


J 


CHAPTER IV. 

SIMPLE CURVES. 

37. The curves most generally in use are circular curves, al¬ 
though parabolic and other curves are sometimes used. Circular 
curves may be classed as Simple, Compound, Reversed, or Spiral. 

A Simple Curve is a circular arc, extending from one tan¬ 
gent to the next. The point where the curve leaves the first 
tangent is called the “PC.,” meaning the point of curvature, 
and the point where the curve joins the second tangent is 
called the “ P. P.,” meaning the point of tangency. The P. C. 
and P. P. are often called the Tangent Points. If the tan¬ 
gents be produced, they will meet in a point of intersection 
called the ‘‘Vertex,” V The distance from the vertex to the 
P.C. or P.T. is-called the “Tangent Distance,” T. The dis¬ 
tance from the vertex to the curve (measured towards the 
center) is called the External Distance, E. The line joining 
the middle of the Chord, C, with the middle of the curve sub¬ 
tended by this chord, is called the Middle Ordinate, M. The 
radius of the curve is called the Radius, B. The angle of 
deflection between the tangents is called the Intersection Angle, 
I. The angle at the center subtended by a chord of 100 feet is 
called the Degree of Curve, D. A chord of less than 100 feet 
is called a sub-chord, c; its central angle a sub-angle, d. 

38. The measurements on a curve are made: 

(a) from P. C. by a sub-chord (sometimes a full chord of 
100 ft.) to the next even station, then 

(b) by chords of 100 feet each between even stations, and 
finally, 

(c) from the last station on the curve, by a sub-chord (some¬ 
times a full chord of 100 ft.) to P.T. The total distance from 
P.C. to P. T. measured in this way, is the Length of Curve, L. 

39. The degree of curve is here defined as the angle sub¬ 
tended by a chord of 100 feet, rather than by an arc of 100 feet. 

20 


Simple Curves. 


21 


Either assumption involves the use of approximate methods 
either in calculations or measurements, if the convenient and 
customary methods are followed. It is believed that on the 
merits of the question, it is best to accept the definition given, 
and the practice in this country is largely in harmony with this 
definition. 

Outside of the United States a curve is generally designated 
by its if ad ius. It. In the United States for railroad purposes, 
a curve is generally designated by its Degree, D. 


40. Problem. Given R. 

Required D. 

O 

o 

H 

41. Problem. Given D. 

Required R. 



R sin | D 


100 

2 


sin \ D 


50 

R 


R = 


50 

sin | D 


Example. Given D = 1°. 

P _ 50 50 log 1.698970 

1 ~ sin \D 0° 30' log sin 7.940842 
Ry = 5729.6 log 3.758128 


( 1 ) 

( 2 ) 


42. Problem. 


i?, = 


Given Ry (radius of 1° curve) or Dy. 

Required R a (radius of any given curve of 
degree = D a ). 

50 ™ 50 


sin i Dy 


Ra = 


( 3 ) 


sin \ D a 

R a _ sin | D\ 

Ry sin \ D a 

In the case of small angles, the angles are proportional to the 
sines (approximately), 


r = l D l 
sin | D a 


ID< 


Dy 


R a = Ryf£=R a = R i 

2 a 

But Ry = 5730 to nearest foot, 

5730 


Ra = 


D a 


(approx.) 


(3,1) 


( 4 ) 


Example. 


R l0 = 573.7 by (3), or by Allen’s Table I- 
= 573.0 by (4) (approx.) 














22 Railroad Curves and Earthwork. 

5 

Some engineers use shorter chords for sharp curves, as 1° to 

7°, 100 ft.; 8° to 15°, 50 ft.; 10° to 20°, 25 ft. R a = ^9 i s 
then very closely approximate. ® a 

Values of R and D are readily convertible. Table I., 
Allen, serves this purpose, giving accurate results or values. 
In problems later, where either R or D is given, both will, in 
general, be assumed to be given. Approximate values can be 
found without tables by (4). The radius of a 1° curve = 5730 
should be remembered. 

43. Problem. Given I, also R or D. 

Required T. 

AOB = NVB = I 
AO = OB — R 
AV = VB =.T 
T = R tan \ I (5) 

Example. Given D = 9 ; 1=. 00° 48'. 
Required To. 

Table I., i?c>log = 2.804327 
30° 24' log tan = 9.768414 

To — 373.9 log 2.572741 



44. Approximate Method. 


Ti = Ri tan ' / • T a = R a tan J / 


Ti fii D a D a 


(approx.) 


T a = 


T\ 

D a 


(approx.) 


( 0 ) 


Table III., Allen, gives values of T\ for various values of I. 
4 able IV., Allen, gives a correction to be added after divid¬ 
ing by D a . 






Simple Curves. 


Example. As before. Given D = 9 ; 7 = 60° 48 r . 

4 

Required Tg. 

Table III., T x 60° 48' = 3361.6(9 

Tg = 373.51 (approx.) 

Table IV., correction, 9° and 61° = .38 

Tg = 373.9 (exact) 

the same as before 

45. Problem. Given 7, also R or D. 

Required E. 

Using previous figure, 

VH = E = R exsec \ I 

Table XIX. gives natural exsec. 

Table XV. gives logarithmic exsec. 

Approximate Method. 
method used for (6), 

E a = ~ (approx.) 

■Da 

Table V. gives values for E x . 

46. Problem. Given 7, also R or D. 

Required M. 

FH = M = R vers \ I 

Table XIX. gives natural vers. 

Table XV. gives logarithmic vers. 

Table II. gives certain middle ordinates. 

47. Problem. Given 7, also R or D. 

Required chord AB = C. 

C = 2 R sin i 7 

Table VIIL gives values for certain long chords. 




24 Railroad Curves and Earthwork. 

48. Transposing, we find additional formulas, as follows : 


from (5) 

R = T cot \ I 

(li) 

(7) 

B = E 

exsec ^ I 

(12) 

(9) 

II 

a ^ 

(13) 

(10). 

R- C 

2 sin \ I 

(14) 

(4) 

D a = (approx.) 

li a 

(15) 

(0) 

D a = (approx.) 

1 a 

(16) 


(8) D a =^- (approx.) (17) 

l^a 

49. Problem. Given sub-angle cl , also li or D. 

JRequired sub-chord c. 


c = 2 J? sin \ d 


(18) 


Approximate Method. 


100 = 21? sin-JD 


sin \d d , v 

° = —(approx.) 


100 sin | D D 


( 10 ) 


The precise formula is seldom if ever used. 


50. Problem. 


Given sub-chord c, also R or D. 
Required sub-angle d. 


d = 

d 
2 


cD 

100 


( 20 ) 


c D 
100 2 


( 21 ) 









Simple Carves. 


25 


A modification of this formula is as follows : 

d _ cD 
2 ~ 200 

for 1) = 1 


d 00' n or 
- = c — = cx 0.3' 
2 200 


for any value D a 

f = c x 0.3' x D a (value in minutes) (22) 
2 

This gives a very simple and rapid method of finding the 

value of - in minutes, and the formula should be remembered. 
2 


Example. 


Given sub-chord = 63.7. D — 0° 30'. 


acquired sub-angle d 



I. By (20) 63.7 

6.5 = D 
3185 
3822 
414.05 

4°. 14 
00 ' 

d = 4° 08' ’ 


d _ 2 o 04 / 

2 


II. By (21) 63.7 
3.25 
3185 
1274 
1911 


D 


207.025 


2.07 

_6(V 

^ = 2° 04' 
2 


III. By (22) 63.7 

0.3 
19.11 
6.5 


9555 

11466 

124.215 minutes 


( 1 = 2° 04' 
2 


Method III. seems preferable to I. or II. 












26 


Railroad Curves and Earthwork. 


51. Problem. Given I and D. 

„'Required L. 

(a) When the P. C. is at an even station, D will be contained 
in / a certain number of times n, and there will remain a sub¬ 
angle d subtended by its chord c. 


/ _ i d_ _ 
D U + 1) 


n -1-(approx.) 

100 ' 


100 — = 
D 


= 100 n + c = L (approx.) 


(&) When the P.C. is at a sub-station the same reasoning 
holds, and 


Transposing, 


L = 100 ~ (approx.) 

(23) 

I = (approx.) 

100 ' 

(24) 

D _ 100 z 

L 

(25) 


These formulas (23) (24) (25), though approximate, are the 
formulas in common use. 

Example. Given 7° curve. I = 39° 37'. 

Required L. 

I = 39° 37 
D = 7 )39.0107° 

5.6595+ 

L = 566.0 

: <? . 

Example. Given D and L. 

Required I. 

Given 8° curve < 


also, P. T. — 93 + 70.1 
P.C. = 80 + 49,3 
L- 7 20.8 

D = _8 

57.004 
_ 00 ' 

- 39.84 


I = 57° 40' 









Simple Curves. 


27 


52. Method of Deflection Angles. 

If at any point on an existing curve a tangent to the curve be 
taken, the angles from the tangent to any given points on the 
curve may be measured, and the angles thus found may be 
called Total Deflections to those points (as NAl, NA2, NA3). 

In laying out successive points upon a straight line (as on a 
“ Tangent ”), each point is generally fixed by 
(a) measurement from the 
preceding point and 
(ft) line; 

the line on a tangent will 
be the same for all points. 

Similarly, in laying out 
a curve, successive points 
may be fixed by 
(a) measurement from the 
preceding point and 
(ft) line; 

the line in this case, for the curve, will be that found by using 
the total deflection calculated for each point. In the figure pre¬ 
ceding, the chord distance A I and the total deflection NAl fix 
point I ; the chord distance 1-2 and total deflection NA2 fix 
point 2 ; and 2-3 and NA3 fix 3. A curve can be conveniently 
laid out by this method if the proper total deflections can be 
readily computed. 

53. Problem. Given a Parabolic Curve. 

Required total deflections to 

B-C-D and chord lengths 
AB - BC - CD. 

Give results to the nearest 
minute or nearest foot. 

54. Simple Curves. 

In the case of “ Simple Curves ,” the “total deflections” can 
be readily computed, and the method of “deflection angles” is 
therefore well adapted to laying them out. 



See Chapter VII for 
equation of parabola. 





28 


Railroad Curves and Earthwork. 

55. Problem. To find the Total Deflections for a Simple 
Curve having given the Degree. 

I. When the curve begins and ends at even stations. 

The distance from station to station is 100 feet. The deflec¬ 
tion angles are required. 


V 



An acute angle between a tangent and a chord is equal to 
one half the central angle subtended by that chord 

A I = 100 VAI=£Z> 

The acute angle between two chords having their vertices in 
the circumference is equal to one half the arc included between 
those chords. 

1 — 2 = 100 and I A 2 — \D Similarly, 

2 - 3 = 100 and 2 A 3 -\D 

3 - B = 100 and 3 A B = \ D 

This angle \D is called by Henck and Searles the Deflection 
Angle, and will be so called here. Shunk and Trautwine call 
it the “ Tangential Angle.' , ' > The weight of engineering opinion 
appears to be largely in favor of the “ Deflection Angle.” 

The “ Total Deflections' 1 ' 1 will be as follows: 

V A I = \D 

V A 2 = V A I +\D 

VA3 = VA2 + {D 

VAB will be found by successive increments of \ D. 

VAB = VBA = \1. This furnishes a “ check ” on the compu¬ 
tation. 

II When the curve begins and ends with a sub-chord. 

V A I = \d . 

V A2 = VA I + i D 

VA3 = VA2 + \D 



Simple Curves . 


29 


VAB is found by adding | d., to previous “ total deflection.” 

VAB = VBA — \ I. This furnishes “ check.” The total deflec¬ 
tions should be calculated by successive increments; the final 
“check” upon \I then checks all the intermediate total 
deflections. The example on next page will illustrate this. 

56. Field-work of laying out a simple curve having given the 
position and station of P. C. and P. T. 

( а ) Set the transit at P.C. ^A). 

(б) Set the vernier at 0. 

(c) Set cross hairs on V (or on N and reverse). 

( d ) Set off | d\ (sometimes \D) for point I. 

(e) Measure distance Ci (sometimes 100) and fix I. 

(/) Set off total deflection for point 2. 

(g) Measure distance 1-2 = 100 and fix 2, etc. 

( h ) When total deflection to B is figured, see that it = \ 7, 
thus “checking” calculations. 

(i ) See that the proper calculated distance c 2 and the total 
deflection to B agree with the actual measurements on the 
ground, checking the field-work. 

(&) Move transit to P. T. (B). 

( l ) Turn vernier back to 0, and beyond 0 to \I. 

(tn ) Sight on A. 

( n ) Turn vernier to 0- 

(o) Sight towards V (or reverse and sight towards P), and see 
that the line checks on V or P. 

It should be observed that three “checks” on the work are 
obtained. 

1. The calculation of the total deflections is checked if total 
deflection to B = \ I. 

2. The chaining is checked if the final sub-chord measured 
on the ground = calculated distance. 

3. The transit work is checked if the total deflection to B 
brings the line accurately on B, 

The check in I is effective only when the total deflection for 
each point is found by adding the proper angle to that for the 
preceding point. 

The check in 3 assures the general accuracy of the transit 
work, but does not prevent an error in laying off the total 
deflection at an intermediate point on the curve. 


30 


Railroad Curves and Earthwork. 


57. Example. Given Notes of Curve 

P. T. 13 + 45.0 
P.C. 10+74.0 


6° curve L 


Bequired the “total defections ” 
to sta. 11 Ci = 20 

7.8 
_6° 

— = 46.8 = 0° 47' to 11 

2 3° 

c 2 = 45 3° 47' to 12 

.3 3^_ 

13.5 6° 47' to 13 

_6° 

^ = 81.0 = 1° 21' 

2 8° 08' to 13 + 45 

13 + 45.0 
10 + J74.0 
2 71.0 = L 

_6° 

10.20 16° 

60' 

15.6' _16| 

16° 16' = I 
8°08' = a / “check” 


58. Caution. 

If a curve of nearly 180° = I is to be laid out from A, it is 
evident that it would be difficult or impossible to set the last 
point accurately, as the “intersection” would be bad. It is 
undesirable to use a total deflection greater than 30°. 

It may be impossible to see the entire curve from the P.C. 
at A. 

It will, therefore, frequently happen that from one cause or 
another the entire curve cannot be laid out from the P.C., and 
it will be necessary to use a modification of the method de¬ 
scribed above. 












Simple Curves. 


31 


59. Field-work. When the entire curve cannot he laid out 
from the P.C. 

First Method. 

(а) Lay out curve as far as C, as before. 

(б) Set transit point at some convenient point, as C (even 
station preferably). 

(c) Move transit to C. 

(d) Turn vernier back to 0, and beyond 0 by the value of 
angle VAC. 

(e) Sight on A. 

(/) Turn vernier to 0. See that transit line is on auxiliary 
tangent NCM (VAC = NCA being measured by a arc AC). 

(. g ) Set off new deflection angle (±d or \ Z>). 

( h ) Set point 4, and proceed as in ordinary cases. 


v 



60. Second Method. 

(a) Set point C as before, and move transit to C. 

(b) Set vernier at 0. 

(c) Sight on A. 

(d) Set off the proper ‘ 4 total deflection ’’ for the point 4=VA 4 

NCA 4- MC 4 = VA 4, each measured by \ arc AC 4. 

(e) Reverse transit and set point 4. 

(/) Set off and use the proper “total deflections” for the 
remaining points. 

The second method is in some respects more simple, as the 
notes and calculations, and also setting off angles, are the same 
as if no additional setting were made. By the first method the 
deflection angles to be laid off will, in general, be even minutes, 




32 


Railroad Carves and Earthwork. 


often degrees or half degrees, and are thus easier to lay off. It 
is a matter of personal choice which of the two methods shall 
be used. It will be disastrous to attempt an incorrect combina¬ 
tion of parts of the two methods. 

61. Field-work of finding P.C. and P.T. 

In running in the line, it is common and considered advisable 
to establish “F,” determine the station of “F,” and measure 
the angle I. Having given 1 only, an infinite number of curves 
could be used. It is, therefore, necessary to assume additional 
data to determine what curve to use. It is common to proceed 
as follows : 

(a) Assume either (1) D directly. 

(2) E and calculate D. 

(8) T and calculate D. 

It is often difficult to determine off-hand what degree of curve 
will well fit the ground. Frequently the value of E a can be 
readily determined on the ground. The determination of D 
from E a is readily made, using the approximate formula 

D a = —• Similarly, we may be limited to a given (or ascer- 

E a rrt 

tainable) value of 2’ ft , and from this readily find D a = —• 

T a 

The value of D a adopted will, in general, be taken to the 
nearest \° (perhaps only to nearest degree) rather than at the 
exact value found, as above. (Some engineers use 1° 40' = 100' 
and 3° 20' = 200', etc., rather than 1° 30' or 3° 80', etc.). 

(b) From the data finally adopted T is calculated anew. 

(c) The instrument still being at F, the P. T. is set by laying 
off T. 

(cl) The station of P. C. is calculated and P C. set. 

(e) The length of curve L is calculated, and station of P.T. 
thus determined (not by adding T to station of F). 

'Total deflections should be all calculated and entered in note- 
b< >ok. 

Whether 7), 7?, or T shall be assumed depends upon the 
special requirements in each case. Curves are often run ou& 
from P.C. without finding or using F, but the best engineering 
usage seems to be in favor of setting F, whenever this is at all 
oracticable, and from this finding the P.C. and P.T. 


Simple Curves. 


62. Example. Given a line , as shown in sketch. 

Required a Simple Curve to connect the 
7 an gents. 

P. T. is to be at least 300 ft. from end of line. 

Use smallest degree or half degree consistent with this. 

Eind degree of curve and stations of P.C. and P. T. 


O 


Table III. 22° 14' T x 


T x = 1125.8( 5° 
225.16 

Table IV. corr. _.07 I 

T= 225.2 

L 


It will be noticed that the station of the P. T. is found by 
adding L to the stations of the P. C. ( not by adding T to the 
station of V ). 

Similarly Given E = 17 ft. 

Table V. 22° 14' E x = 109.6/_17 = E 

102 \ 6.4 + use 6° 30' curve 

T x - 1125.8(0.5 70 

173.20 7= 22°.2333( 6.5 V = 46 + 72 7 

corr. _ AC L = 342.1 T = 1 + 73 3 

T = 173.3 * P C. = 44 + 99.4 

L = 3 + 42. 1 
P. T. = 48 + 4 1.5 

Under the conditions prescribed a 1 ove, when T is given, the 
degree, or half degree, next larger must be used, in order to 
secure at least the required distance (to end of line above). 

When E is given, the nearest half degree is generally used. 



300 

= 1125.8 / 230 = T (approx.) 

92 \ 4.0 — use 5° curve 

205 

P = 46 + 72.7 
= 22° 14' T= 2 + 25.2 

= 22°.2333( 5° P.C. - 44 + 47.5 
- 444.7 L = 4 + 44,7 

P. T. = 48 + 92.2 
















34 Railroad Curves and Earthwork. 

S 

63. Form of Transit Book (left-hand page). 


(Date) 

(Names of Party) 


Station 

Points 

114 


113 


112 


III 


NO 


109 

© + 90.0 P. T. 

108 

/ 

107 

106 

0+ 68.0 V 

105 

0+40.0 P.C 

104 


103 


102 


101 


100 


99 


98 



Descrip, of 
Curve 


Total 

Deflect. 


11° 15' 


B = 1146.3 
L = 450.0 
T= 228.0 
7= 22° 30' 
5° Right 


9° 00' 
6 ° 30 ' 
4° 00' 
1° 30' 


Observed 

Course 


N 46° 00' E 


N 23° 15' E 


V is not a point on the curve. Nevertheless, it is customary 
to record the station found by chaining along the tangent. 

The right-hand page is used for survey notes of crossings of 
fences and various similar data. It seems unnecessary to show 
a sample here. 


















Simple Curves. 


35 


64 Metric Curves. 

In Railroad Location under the “Metric System” a chain of 
100 meters is too long, and a chain of 10 meters is too short. 
Some engineers have used the 30-meter chain, some the 25- 
meter chain, but lately the 20-meter chain has been generally 
adopted as the most satisfactory. Under this system a “ Sta¬ 
tion ” is 10 meters. Ordinarily, every second station only is 
set, and these are marked Sta. 0, St a. 2, Sta. 4, etc. On curves, 
chords of 20 meters are used. Usage among engineers varies as 
to what is meant by the Degree of Curve under the metric 
system. There are two distinct systems used, as shown below. 

I. The Degree of Curve is the angle at the center subtended 
by a chord of 1 chain of 20 meters. 

II. The Degree of Curve is the deflection angle for a chord 
of 1 chain of 20 meters (or one half the angle at the center). 

II. Or, very closely, the Degree of Curve is the angle at the 
center subtended by a chord of 10 meters (equal to 1 station 
length). 

For several reasons the latter system is favored here. Tables 
upon this basis have been calculated, giving certain data for 
metric curves. Such tables are to be found in Allen’s Field 
and Office Tables. 

In many countries where the metric system is used, it is not 
customary to use the Degree of Curve , as indicated here. In 
Mexico, where the metric system is adopted as the only legal 
standard, very many of the railroads have been built by com¬ 
panies incorporated in this country, and under the direction of 
engineers trained here. The usage indicated above has been 
the result of these conditions. If the metric system shall in 
the future become the only legal system in the United States, 
as now seems possible, one of the systems outlined above will 
probably prevail. 

In foreign countries where the Degree of Curve is not used, it 
is customary, as previously stated, to designate the curve by 
its radius 72, and to use even figures, as a radius of 1000 feet, 
or 2000 feet, or 1000 meters, or 2000 meters. As the radius is 
seldom measured on the ground, the only convenience in even 
figures is in platting, while there is a constantly recurring incon¬ 
venience in laying off the angles. 


86 > 


Railroad Curves and Earthwork. 


65. Circular Arcs. For general railroad work, the Length of 
Curve is the distance measured by a series of chords as defined 
in § 38 and § 51. For certain purposes the actual length of arc 
is required. Where the line of a street is curved, the length of 
the side line of street, the property line of a lot or estate, may 
be required. This is one instance only. 

Problem. Given the Central angle A and radius B. 

Required the length of arc. 

Table XX. p. 205 Allen gives lengths of circular arcs for 
radius =1. The values for degrees, minutes and seconds are 
added; the sum multiplied by B is the required length of arc. 

Example. A = 18° 43' 29" 18° 0.3141593 

B = 600. 43' 0.0125082 

29" 0,0001406 
0.3268081 

B _ 600 

196.08486 

length of arc = 196.08 

/ 

Where several points are to be set in the curve, the entire 
length may be divided into any desired number of equal parts, 
and the deflection angles determined to correspond. 

If 6 be the desired number of parts, 

6 )196.08 A = 18° 43' 29" 

32.68 | A — 9° 21' 44" = total deflection angle. 

6 )9° 21'44" 

1° 33' 37" 

The deflection angles will be (to nearest \ minute), 

1° 33' 30" 

3° 07' 

4° 41' 

6° 14' 

7° 48' 

9°21'30" 

The length of chord corresponding to 32.68 may also be com¬ 
puted by formula (10). 






Simple Curves. 


37 


66- Problem. Given D and stations of P.C. and P.T. 

Required to lay out the curve by the method 

of Offsets from the Tangent. 



AE' = Ci cos EE' = Ci sin 

EF" = 100 cos FF" == 100 sin a? 

FG" = 100 cos «3 GG" = 100 sin a$ 

FF' = EE' + FF" 

GG' = FF' + GG", etc. 
When AE = 100, then \d becomes \ D. 

For the computations indicated above, always use natural 
sines and cosines. 

For a check, AG' = P sin AOG 

GG' = R vers AOG 
where O is at center of curve. 

If AE, EF, FG are parts of a compound curve, this method is 
still applicable. 

'Phis method of Offsets from the Tangent is a precise method, 
and allows of any degree of precision desirable. 

67. Field-work. 

(a) Calculate AE', E'F', F'G'; also EE', FF', GG' 

(&) Set E', F', G'. 

(c) Set E by distance AE (ci) and EE'. 

(d) Set F “ “ EF (100) and FF'. 

(e) Set G “ “ FG (100) and GG'. 







38 


Railroad Curves and Earthwork. 


68. Problem. Given D and the stations of P. C. and P.T. 

Required to lay out the curve by the method 

of Deflection Distances. 

I. When the curve begins and ends at even stations. 

In the curve AB, let 

AN be a tangent 
AE any chord = c 
EE' perp. to AE' = a = 

‘ 4 tangent deflection ’ y 
FF' = BB' = the 
44 chord deflection ” 

AO = EO = B 

Draw OM perpen¬ 
dicular to AE. 



Then 


EE':AE= ME : EO 


a : c = - : E 
2 


a = 


2 R 


(26) 


FF' = 2 a ; AF' = AE produced 
When AE is a full station of 100 feet, 

100 2 


«ioo 


2P 


Field-work. 

The P. C. and P. T. are assumed to have been set. 

(a) Calculate aioo- 

(&) Set point E distant 100 ft. from A and distant aioo from 
AE' (AE' < 100 ft. ; AE'E = 90°). 

(c) Produce AE to F' (EF' = 100 ft.), and find F distant 
2 aioo from F' (EF = 100 ft.). 

(d) Proceed similarly until B is reached (P.P.). 






Simple Curves. 


39 


(e) At station preceding B ( P.T .) lay off FG' = aioo 
(FG'B = 90°). 

(/) G'B is tangent to the curve at B (P. T.). 

69. For any given length of chord, the “tangent deflec¬ 


tion ” a = 


2 R 


(26) 


for 1° curve 


Again 


cti = ——; for any ° curve a a = —- 
2 RC 2 R a 

1_ 

— = —y"-- = — (approx.) from (3 A) 
d\ 1 D i 

Pi 

da — d\D a 
100 2 


(26 A) 


«ioo = 


2P 


for n chords of 100, = n = n 2 ffioo. 

2 

for 1° curve and 100 ft. chords «ioo = 0.873 = f ft. (nearly) 
for 1° curve a n = f ft 2 (approx.) 

for any ° curve a n =} w 2 A* (approx.) (27) 


70. Problem, (riven two Curves of degree Di and D /. 

Required the offset between the two Curves 
for any number of even stations. 


From (26 A) 
for 2° curve 


CJa — a\JC a 

« 2 = «i 2; also a 3 — ai 3 
c?2 "F et.3 = a\ (2 -f- 3 ) 


= d\ 5 = a$ 


a f — a\ D f &n&ai = A 

ct f — di=ai(D/ — Di ) = a/-,- 

for 1 chord of 100 ft. f (A — A) = a 

for n chords of 100 ft. a n = j n 2 (A — A) 2 


(approx.) (281 








40 Railroad Curves and Earthwork. 

S " ... 

71. Problem. Given D and the stations of P. C. and P. /. 

Bequired to lay out the Curve by Deflection 
Distances. 

II. When the curve begins and ends with a sub-chord. 



ai: aioo = Ci 2 : 100 2 
a f : aioo = C/ 2 : 100 2 


Oi = aioo 
Of = aioo 


100 2 

cl_ 

100 2 


(29) 


In general it is better to use (29) than a,- = —— 

2 B 


72. Example. Given P. T. 20 -f 42 

P. C. 16 + 25 


6° curve R 


Bequired all data necessary to lay out curve by “ Deflec¬ 
tion Distances.' 1 ' 1 

Calculate without Tables. Result to foot. 


Radius 1° curve = 5730( 6 
6° 955 

100 2 


«ioo = 


= 5.24 


2 x 955 
2 aioo — 10.47 

a 7 5 = 0.76 s * x 5.24 =, 2.95 
a 42 = 0.42 2 x 5.24 = 0.92 


1910)10000(5.235 + 
955 
450 
382 


680 

573 

1070 

955 


Table II. gives aioo = 5.234 (precise value) 














Simple Curves. 


41 


The distance AE' is slightly shorter than AE. It is generally 
sufficient to take the point E' by inspection simply. If desired 
for this or any other purpose, a simple approximate solution of 
right triangles is as follows : 


73. Problem. Given the hypotenuse (or base) and altitude. 

Required the difference between base and hy¬ 
potenuse, or in the figure, c — a. 

c 2 - a 2 = A 2 

(c — a) (c + a) = h 2 

A 2 A 2 

c — a — —— = — (approx.' 

c a 2 c 

Wherever A is small in comparison with a or c, the approxi¬ 
mation is good for ordinary purposes. 



Example, c = 100 c — a = — = 0.50 

200 

A = 10 a = 99.50 

The precise formula gives 99.499. 

74. Field-work for Case II., p. 40. 

(а) Calculate $ioo, $/. Remember that tangent deflec¬ 
tions are as the squares of the chords. 

«ioo is found in Table II., Allen, as “tangent offset.” 

(б) Find the point E, distant $ £ from AE' and distant c £ from 
A. (AE'E = 90°.) 

(c) Erect auxiliary tangent at E (lay off AA' = af). 

(d) From auxiliary tangent A'E produced, find point F. 

(FF' = aioo ; EF = 100 ; EF'F = 90°). 

(e) From chord EF produced, find point G. 

(GG' = 2 ccioo ; FG' = FG = 100). 

(/) Similarly, for each full station, use 2 $i 0 o, etc. 

( g ) At last even station on curve, H, erect an auxiliary tan¬ 
gent (lay off GG" = $ioo ; GG"H = 90°). 

(A) From G"H produced, find B (B'B = a/, etc.). 

(i) Find tangent at B (HH" = $/; HH"B = 90°). 

The values of «ioo 5 , «/, should be calculated to the nearest 

ifhr 






42 


Railroad Curves and Earthwork. 


75. Caution. The tangent deflections vary as the squares of 
the chords, not directly as the chords. 

Curves may be laid out by this method without a transit by 
the use of plumb line or “flag” for sighting in points, and with 
fair degree of accuracy. 

For calculating « 10 o, a «/, it is sufficient in most cases to use 

5780 


the approx, value R a = 


D n 


A curve may be thus laid out 


without the use of transit or tables. 

For many approximate purposes it is well and useful to 
remember that the “chord deflection” for 1° curve is 1.75 ft. 
nearly, and for other degrees in direct proportion. A head 
cliainman may thus put himself nearly in line without the aid 
of the transit man. 

The method of “Deflection Distances” is not well adapted 
for common use, but will often be of value in emergencies. 


76. Problem. Given D and stations of P. C. and P. T. 

Required to lay out the curve by “ Deflection 
Distances ” when the first sub-chord is 
small. 

Caution. It will not be satisfactory in this case to produce 
the curve from this short chord. The method to be used can 
best be shown by example. 

Let PC =41 + 90. 


Field-work. 

Method 1. 

(a) Set sta. 42 using c = 10 and « 10 — «ioo 


1Q3 

100 2 ’ 


(5) Set sta. 43 (100 ft. from 42) offsetting « 110 from tangent, 
(c) Set sta. 44 by chord produced and 2 a 100 offset. 

Method 2. 

(a) Set a point on curve produced backwards, using 

90 2 


c = 90 and « 90 = aioo 


100 2 


(b) Set sta. 42, using c = 10 and « 10 as above. 

(c) Set sta. 43 by chord produced and 2 a 10 o offset. 

A slight approximation is involved in each of these methods. 
Method 1 involves less labor. 





Simple Curves. 


43 


77. Ordinates. 

Problem. Given D and two points on a curve. 

Required the Middle Ordinate from the 
chord between those two points. 

By (9), 3I=R\ers^I 

for 100 ft. chord 31= R vers \ D 

between points 200 ft. apart 31 = R vers D. 

Let A = angle at center between the two points. 

31 = R vers \ A. 


78. Problem. Given R and c. 

Required M. 


OL = ^R* _ 

HL = jtf= (31) 


M=li-^R + £j(32) 


H 



Table XXI., Allen, gives squares and square roots for certain 
numbers. If the numbers to be squared can be found in this 
table, use (31). Otherwise use logarithms and (32). 


79. Problem. Given R and C. 

Required the Ordinate at any given point Q. 

Measure LQ = q. Then KN — y/R* — q l 

Lo =VMi)' 


KQ = KN - LO =■ V(J S + q) (It - q ) ~V(' K + i) [ B ~§) ( ' 3S - > 












44 


Railroad Curves and Earthwork . 


80. When C = 100 ft. or less, an 
approximate formula will generally 
suffice. 

Problem. Given It and c. 

Required M (approx.) 

HL : AH = — : R 
2 



O 


31 = 


AH' 2 
2 R' 


Where AB is small compared with R , 


AH = - (approx.) 
2 



J— 

8 R 


(approx.) 


(34) 


81. Example. Given C = 100, D = 0°. 

Required M. 

R 9 = — = 636.7 
9 8 

5093.6)10000.(1.963 = 31 
50936 
490640 

Precise value 458424 

31 = 1.965 332160 

305616 

16544 


Table XXVII., Allen, gives middle ordinates for curving rails 
of certain lengths. 


82. Problem. Given R and c. 

Required Ordinate at any given point Q 
Approximate 3Iethod. 


I. Measure LQ = q 


31= HL = 



KK' 


HK 2 
2 R 



B 















Simple Curves. 


45 


Since HK = q (approx.) KK' = M (approx.) (35) 

(I) 

KQ = M- KK' 

When \ as in fi S llre > KK' = ~ and KQ = ? M (approx.) 

2 

When = - VW = jjj M (approx.) 

2 

When ^ TU = ^ ilf (approx.) 

2 

The curve thus found is accurately a parabola, but for short 
distances this practically coincides with a circle. 


83. II. Approximate Method. Measure LQ and QB 

(?r 


M = 

KQ = 
KQ = 


2 B 


KK' = (approx.) 


(i ) 2 ~ 32 (f +? )(H) 


2 B 


2 B 

AQ x QB 
2 B 


(approx.) 


(approx.) 


(36) 


Sometimes one, sometimes the other of these methods will be 
preferable. 


84. Example. Given C = 100, D = 9°. 

M = 1.965 from Tables. 


Bequired , Ordinate at point 30 ft. distant 
from center toward end of chord. 


I. 30 ft. = — x - 
50 2 

KK' = — x 1.965 
25 9 

25 )17.685 

.70740 

M~ 1.965 

Ordinate = 1.258 


II. AQ = 80 

BQ = 20 
1273.4)1600(1.256 
1273.4 

Bi = 6730. 3266 O 

B<i = 636.7 25468 

2 i? 9 = 1273.4 71920 

63670 


Precise result for data above = 1.260. 


8250 















46 Railroad Curves and Earthivork. 

S 

85. Problem. Given R and c. 

Required a series of points on the curve . 



H 



M= HL = 


8 R 


(approx.) 


AH 2 

RS = (approx.) 

O li 


AH = - (approx.) 
2 


„ 4 M, 

RS = — - T (approx.) 

DC 

PN = — (approx.), etc., as far as desirable. 

This method is useful for many general purposes, for ordi¬ 
nates in bending rails among others. 


86. Problem. Given a Simple Curve joining two tangents . 

Required the P. C. of a new curve of the same 
radius which shall end in a parallel tan¬ 
gent. 


Let AB be the given curve. 


A'B' “ “ required curve. 

B'E — p — perpendicular distance between tangents. 


Join BB'. 

Then A A' = 00' = BB' 


Also B'BE = V'VB = 1 
BB' sin I = p 

BB' = AA' = -4L- (37) 

sin / 

When the proposed tangent 
is outside the original tangent, 
the distance AA is to be added 



to the station of the P. C. When inside, it is to be subtracted. 










47 


Simple Curves. 


87. Problem. Given a Simple Curve joining two tangents. 

Required the Radius of a new curve which 
with the same P. C. shall end in a parallel 
tangent. 



Let AB be the given curve of 
radius R = AO. 

B'E = p = perpendicular 
distance. 

AB' the required curve, 
radius = R'. 

Draw chords AB, AB'; 
also line BB'; 
also BL parallel to AO'. 


Then 


BL = 00' 

= R' -R 
= B'L 

BLB' = AO'B' 


= I 

BLvers BLB' = B'E 
(.R' — R) vers I = p 


(R 1 -R) = 


P 

vers I 


(38) 


Since VAB = V'AB', AB and AB' are in the same straight line. 
BB'= 2 BL sin \ BLB' 

= 2(22' -22 )siniJ ( 39 ) 


When the proposed tangent is outside the original tangent (as 
it is shown in the figure), the above formula applies, and 

R’>R. 

When the proposed tangent is inside the original tangent, the 
formula becomes 

R - RJ = -V— 
vers I 



and R r < R. 






48 


Railroad Curves and Earthwork. 


88. Problem. Given a Simple Curve joining two tangents. 

Required the radius and P.C. of a new curve 
to end in a parallel tangent with the new 
P. T. directly opposite the old P. T. 

Let AB be the given curve of 
radius = R. 

A'B' the required curve of 
radius R'. 

BB' = p. 

Draw perpendicular O'N 
and arc NM 

Then O'M = B'M - B'O' 

= B'M - BM = BB' 
O'M =p 

ON exsec NOO' = O'M 

(P-P')exsec I =p ; R - R' = -- 1 - (41) 

AA' = O'N = ON tan NOO' 

AA' — {R — R') tan I (42) 

When the new tangent is outside the original tangent (as in 
the figure), P > R' and AA' is added to the station of the P. C. 

When the new tangent is inside the original tangent, R < R\ 

R' — R = —^—, and AA' is subtracted from station of P. C. 

exsec I 

89 Problem. To find the 
Simple Curve that shall join 
two given tangents and pass 
through a given point. 

With the transit at V, the 
given point K can often be 
best fixed by angle BVK and 
distance VK. If the point K 
be fixed by other measure¬ 
ments, these generally can 
readily be reduced to the 
angle BVK and distance VK. 













Simple Curves. 


49 


90. Problem. Given the two tangents intersecting at F, the 

angle /, and the point K fixed by angle 
BVK = /3 and distance VK = b. 

Required the radius R of curve to join the 
two tangents and pass through K. 

In the triangle VOK we have given 

VK = b and OVK = - 1 - 8 . 0 ~ 1 - p 

2 

Further VO = B OK = R 

cos \ I 

VO :0K = sin VKO : sin OVK 


cos £ I 


From data thus found, the triangle VOK may be solved for R. 
In solving this triangle the angle VOK is often very small. A 
slight error in the value of this small angle may occasion a 
large error in the value of R. In this case use the following 
Second Method of finding R after VOK has been found. 


= sin VKO : cosQ/-f /3) 


sin 


VK0 = cos (i 7 + ^ 

cos \ I 


(43) 


Find AOK = \ I + VOK Also DVK = I + p 

Then R vers AOK = LK 

= b sin DVK 

jy _ b sin DVK 
vers AOK 


(44) 


91. Problem. Given R, I. /3(BVK). 

Required b (VK). 

In the triangle VOK 

OK = R; 0V = — 7? — 
cos \ 1 

OVK = 90 —(£7+ Pj 
Solve triangle for b. 

Also find VOK and station of K if desired. 







50 


Railroad Curves and Earthwork. 

j 



92. Problem. To find the point 
where a straight line intersects a 
curve between stations. 

Find where the straight line V'K 
cuts VB at V'. 

Measure KV'B. 

Use V' as an auxiliary vertex. 

Find I' from V'B by (5). 

Solve by preceding problem. 


93. Approximate Method. 

Set the middle point H by method of ordinates. , 

If the arc HB is sensibly a straight line, find the intersection 
of HB and CD. 

Otherwise set the point G by method of ordinates, and get 
intersection of HG and CD. 



. Additional points on the arc may be set if necessary, and the 
process continued until the required precision is secured. 

The points H and G can be set without the use of a transit 
with sufficient accuracy for many purposes, a plumb line or flag 
being used in “sighting in.” 


94. Problem. Given a Simple Curve and a point outside 

the curve. 

Required a tangent to the curve from that 
point. 



Let BDE be the given 
curve. 

P the point out¬ 
side the curve. 

BL a tangent at B. 
Measure LBP, also BP. 








Simple Curves. 


51 


In the triangle BPO we have given PBO, BP, BO. 

Solve the triangle for BOP and OP. 

Then cos DOP = — = A 

OP OP 

BOD = BOP - DOP 

From BOD find station of D from known point B. 

It should be noted that if log OP is found, this can be used 
again without looking out the number for OP. Other similar 
cases will occur elsewhere in calculation. 

When for any reason it is difficult or inconvenient to measure 
BP directly, the angles CBP, BCP and the distance BC may be 
measured and BP calculated. 

95. Approximate Method. 

Field-work. 

(а) From the station (B) nearest to the required point D, 
find by the approximate method where BP cuts the curve at C. 
(If E be the nearest station, produce PC to B.) 

(б) Assume D with BD slightly greater than CD, and with 
transit at P. C. set the point D (transit point) truly on the curve. 

(c) Move the transit to D, and lay off a tangent to the curve 
at D. This will very nearly strike P. 



(d) If the tangent strikes away from P, at Q, measure QDP, 
and move the point D (ahead or back as the case may be) a dis¬ 
tance c due to an angle at the center d = QDP. The tangent 
from this new point ought to strike P almost exactly. 

In a large number of cases the point D will be found on the 
first attempt sufficiently close for the required purpose. 

If a tangent between two curves is required, similar methods 
by approximation will be found available. 







52 


Railroad Curves and Earthwork . 

Obstacles. 

When any obstacle occurs upon a “ tangent,” the ordinary 
methods of surveying for passing such obstacle will be used. 

96. When V is inaccessible. 

Measure VLM, VML, LM. 

I = VLM + VML 

LV and VM are readily calculated, 
and AL and MB determined. 

In some cases the best way is to 
assume the position of P. O. and run 
out the curve as a trial line, and 
finally find the position of P. C. cor¬ 
rectly by the method of formula (37). 



97. When the P.C. is inaccessible. 


Establish some point D (ap even 
station is preferable) by method of 
“offsets from Tangent” or otherwise. 

Move transit to B (P. T .), and run 
out curve starting from D and checking 
on tangent VB. 


98. When the P. T. is inaccessible. 




With instrument still at P, set some 
convenient point D, move transit to 
PC., and run in curve to D, and then 
pass the obstacle at B as any obstacle 
on a tangent would be passed. 








Simple Curves. 


53 


99. When Obstacles on the Curve occur so as to prevent 
running in the curve, no general rules can well be given. 
Sometimes resetting the transit in the curve will serve. Some¬ 
times, if one or two points only are invisible from the transit, 
these can be set by “ deflection distances ,” and the curve con¬ 
tinued by “ deflection angles ,” without resetting the transit. 
Sometimes “ offsets from the tangent ” can be used to advan¬ 
tage. Sometimes points can be set by “ ordinates ” from 
chords. Sometimes the method shown on page 48, § 92, assum¬ 
ing an auxiliary V, is the only one possible. 

It should be borne in mind that it is seldom necessary that 
the even stations should be set. If it be possible to set any 
points whose stations are known and which are not too far 
apart, this is generally sufficient. 

Finally, for passing obstacles and for solving many problems 
which occasionally occur, it is necessary to understand the 
various methods of laying out curves, and to be familiar with 
the mathematics of curves ; and, in addition, to exercise a rea¬ 
sonable amount of ingenuity in the application of the knowledge 
possessed. 


I 


CHAPTER V. 


COMPOUND CURVES. 

100. When one curve joins another, the two curves having 
a common tangent at the point of junction, and lying upon the 
same side of the common tangent, the two curves form a Com¬ 
pound Curve. 

When two such curves lie upon opposite sides of the common 
tangent, the two curves then form a Reversed Curve. 

In a compound curve, the point at the common tangent where 
the two curves join, is called the P.C.C. , meaning the “point 
of compound curvature.” 

In a reversed curve, the point where the curves join is called 
the P.R. C., meaning the “point of reversed curvature.” 

101. Field-work. 

Laying out a compound curve or a reversed curve. 

(а) Set up transit at P. C. 

(б) Run in simple curve to P. C. C. or P.R. C. 

(c) Move transit to P. C.C. or P.R. C. 

(d) Set line of sight on common tangent with vernier at 0. 

(e) Run out second curve as a simple curve. 

It is not desirable to attempt to lay out the second curve 
with the transit at the P. C. It is not a simple and convenient 
process to calculate the total deflections from the P. C. to a 
series of points on the second curve. It may readily be shown 
that adding the chord deflection for the second curve to the 
total deflection for the P.C.C. or P.R.C. will yield an incor¬ 
rect result. Resetting at the P. C. C. or P.R.C. is quite simple, 
and the process of running in the second curve is similar in 
principle to that of § 59, page 31. 

54 


Compound Curves . 


55 


.102. Data Used in Compound Curve Formulas. 



103. Problem. Given R t , E s , I s . 

Required 7, TT s . 



Draw the common tangent DCE. 
Then 7 = 7 ( -f £ 

AD = CD = Ei tan | 

EB = CE = E s tan a 7, 

or find CD and CE using Allen’s Ta¬ 
ble III. and the correction, Table IV. 

In the triangle DVE we have one 
side and three angles 

DE = Ri tan \ /* + R„ tan \ R (45) 
VDE = Ii 


VED = /. 


Solve for VD and VE 


DVE = 180-7 


AV = AD + VD = Ti 
VB = BE + VE = T t 


This problem will be of use in making calculations for platting, 
in cases where the curves have been run in without finding V 
on the ground. The points D and E will seldom be fixed on the 
ground, and in cases where V is set, the problem is likely to 
take some form shown in one of the problems following. 








56 


Railroad Curves and Earthwork 


104. Problem. Given T s , R,, R ir I. 

Required Ti, 7j, 1$. 

A l v u 



Draw arcs NP and KC. 

Draw perpendiculars MP, LP, SB, UB. 

Then AM = LP 

AN = R t = KP 

NM = LK = LS - KS 
OP vers NOP = VB sin VBS — PB vers KPB 
{Ri — R t ) vers 7 { = J* sin I — R s vers 7 

vers Ii = 


T s sin I — R s vers 7 


Ri-R . 

I. = I-h 

AV= MP + SB - UV 
Ti = ( R t — R s ) sin h + R s sin I — T, cos 7 

105. Problem. Given T s , R s , 7 g , 7. 

Required 7), R iy I t . 

Ii = 1—1, 

T) -n — Ts sin I — R g vers 7 

—-=- 

vers Ii 

Ti — ( R t — R s ) sin I t + R s sin I— T, cos 7 

106. Problem. Given T t , 1\ , 7? s , 7. 

Required R t , I\ , 7,. 

T s sin 7 — R s vers 7 


(40) 


(47) 


(48) 

(49) 


tan \Ii = 

2 77+ 77 cos 7-7? g sin 7 

R - R = Tl + Tg cos 1 ~ sin 1 

sin 7 t 


(50) 


( 51 ) 











Compound Curves. 


57 



107. Problem. 

Given 1\, It,, B s , I. 
Bequired T s , 7*, 

Draw arcs NP, KC. 

Draw perpendiculars OK, 
AS, PM, VU. 


= KN 

MN = LM - LN 
= KN — LN 
= KL 

LK = MN= KS - LS 
OP vers NOP = AO vers AOK — AV sin VAS 
{Bi — B s ) vers 7, = B t vers I — 77 sin I 

vere 4 = B, vers 7- r, 8 inj 
Bi — B s 

h = I-Is 

VB = AS - PM - AU 

T s = Bi sin I — ( B l — B s ) sin I s — T t cos 7 (53) 

108. Problem. Given 77, B t , I. 

Bequired T a , B s , 7 S . 

Is = I-h 

B, - B. = J? ‘ vereJ - 7 ’- sinJ (54) 

vers 7* 

T s = B t sin I-(Bi- B 6 ) sin I s - 77 cos 7 (55) 

109. Problem. Crwere 77, 77, 7?(, 7. 

Bequired B s , I t , I s . 

Bi vers 7—77 sin 7 


tan | / 8 = 
7?( - B s = 


717 sin 7 — 7) cos 7—77 
Bi sin 7 — 77 cos 7 — 77 


(56) 


sin 7. 


( 57 ) 








58 Railroad Curves and EartJnvork. 

110. Problem. Given, in the figure, AB, VAB, VBA, B s . 

llequired B t , I lf I s , I. 

v Draw arc NP ; also perpendicular 



OP = lh - li t = 


MP 
sin /j 


(59) 


111. Problem. Given, in the figure, AB, VAB, VBA, B\. 

Itequired B s , I t , I,, I. 

Find I, and show that , 

tan i I. = -frvefsJ -ABBmVBA. (60 ) 


— JSj — 


Bi sin I — AB cosVBA 
B t sin 7 — AB cos VBA 
sin L 


(61) 


112. Problem. Given a Simple Curve ending in a given 

tangent. 

A second curve of given radius is to leave this and end in 
a parallel tangent. 

Bequ ired the P. C. C. 

Let AB be the given curve of ra¬ 
dius B t . 

C be the P. C. C. 

CB' the second curve of ra¬ 
dius B s . 

BE = p=distance between tan¬ 
gents. 

Then vers COB = 



B, - lh 


1.62) 











Compound Curves. 


59 


It may sometimes be more convenient or quicker to run in 
a simple curve first and change to a compound curve by the 
method of this problem, rather than to run in the compound 
curve at first. When it is impossible or inconvenient to run in 
the curve as far as B ( P.T. ), the point of intersection D between 
the curve and the tangent may be found, the angle LDN meas¬ 
ured, and BE calculated. In the figure below 

BE = DO vers DOB 

p = Hi vers LDN (63) 


113. Example. Given Notes of Curve. 


22 + 20 P.C. 5° curve B 



Proposed tangent intersects 
curve at 26 + 90. 

Angle between tangent and 
curve = 10 ° 20 '. 

Bequired Station of P.C.C. to 
join proposed tangent , using 
a 7° curve. 

p = vers 10 ° 20 ' 

i ? 5 log 3.059290 
10 ° 20 ' vers 8.210028 

p log 1.269318 


vers COB = ^ p 

As — A 7 

Bs = 

b 7 = 

1146.28 

819.02 

p log 1.269318 
log 2.514893 



327.26 

26 + 90 

1 + 81.3 


19° 24' 

10 ° 20 ' 

vers 8.754425 

25 + 08.7 P. C. C. 


9° 04' 

= 9.0667(5° 


181.3 










60 


Railroad Curves and Earthwork. 


114. Problem. 

gent. 


Then B'E = 


Given a Compound Curve ending in a tan- 

Bequired to change the P. C. C. so as to end 
in a given parallel tangent, the radii re¬ 
maining unchanged. 


T. When the new tangent lies 
outside the old tangent, and the 
curve ends with curve of larger 
radius. 

Let ACB be the given com¬ 
pound curve. 

AC'B' the required curve. 
Produce arc AC to B". 

Draw OB" parallel to PB, and 
B B"E' perpendicular to PB. 

Let B'E = p = perpendicular 
distance between tangents. 

B'E' - BE" 

B'E =(i? t - B g ) versqP'B' —(7?* - B s ) vers COB" 
p =(Bi — B„) vers 7 t ' - (Bi - B t ) vers /, 

P 



vers 7/ = vers I t -f 
AOC' = 7- 7,' 


Bi - B, 


(64) 


II. When the new tangent lies inside the old tangent , and 
the curve ends with the curve of larger radius. 


vers 7/ = vers I t — 


P 


Bi — B t 


(65) 


ITT. When the new tangent lies outside the old tangent, and 
the curve ends with curve of smaller radius. 


vers 7/ = vers 7, — 


Bi — B t 


( 66 ) 


IV. When the new tangent lies inside the old tangent, and 
the curve ends with curve of smaller radius. 

P 


Bi — Bt 


vers 7/ = vers 7 t + 


(67) 












Compound Curves. 


61 


115. Problem. Given a Simple Curve joining two tangents. 


v 



Then 7= 

In the triangle POQ, 

PQ = B t - B t 
QO = Bi — B e 

There are then Given 7, B c . 


Bequired to substitute a sym¬ 
metrical curve with flattened 
ends , using the same P.C. 
and P. T. 

Let AHB be the simple curve of 
radius B c . 

ACDB the required curve in 
which 

BQ = AS =B, 

PC = B, 

ASC = BQD = 7, 

CPD = 7. 

« + 2 7, 

PQO = 7, 

POQ = 180 — \I 
OPQ = \I» 

Bequired Bi , B s , 7,, 7,. 


We may assume any two of the latter (except 7, and 7,), 
and readily calculate the others. 


I. Assume B\ and 7,. 


7, = 7-27, 

PQ : QO = sin POQ : sin OPQ 

B t — B,: Bi - B e = sin \ I : sin \ I„ 

n n _(B t - 7? c )sini7 
Bi — B s — - . . - 


II. Assume Bi and B*. 


sin 17, 


(Bi - B c ) sin 17 
Bi — B, 


( 68 ) 


(69) 






G2 


Railroad Curves and Earthwork. 


III. Assume E s and I e . 


From^68) 


Hi — E s — 


(Ei - E c ) sin \ I 

-- • 

sin £ l s 


E l sin % I s — E s sin \ I g z=. E t sin J / — E c sin J /. 


Ei sin 


11 — Ei, sin l I s = 
Ei = 


E c sin \ I — Eg sin \ I s . 

E c sin i I — Eg sin \ I g 
sin \ I — sin \ I s 


(70) 


116. Problem. Given a Simple Curve joining two tangents. 

Eequired to substitute & curve with flattened 
ends to pass through the same middle 
point. 


Let AB be the given simple curve, 
and H the middle point. 

Erect an auxiliary tangent V , HV ,/ at H. 
The auxiliary intersection angles at 
V' and V" are readily calculated; also 
V'H and V"H. 

Sufficient additional data can be as¬ 
sumed, and the problem solved as a 
problem in compound curves. 

It is not necessary that the curves on the two sides of H 
should be symmetrical. 



Having given V'H = T t 

VV'H = I 


Assume lih>AO 

E, < AO 

and apply (46) and (47). 

Other assumptions of similar sort will allow the use of other 
formulas on pages 56 and 57. 

Both this and the preceding problem will be found of con¬ 
siderable value in revising the alignment of track, and intro¬ 
ducing flatter ends for the curves so that the transition from 
tangent to curve shall be less abrupt. 






Compound Curves. 


63 



117. Problem. Given two simple curves with connecting 

tangent. 

Bequired to substitute a simple curve of 
given radius to connect the two. 

Let DC = l = the given tangent, connecting the two curves 
AD and CB, of radii B s and B h respectively. 


Let EF be required curve of radius B c . 

Join OP, and draw perpendicular OL. 

Then tan LOP = ^ 

U L l 

op = — 1 — 

cos LOP 

In the triangle OPQ we have given 

OP =- l - -; OQ = B c - B s ; QP = B C - B t 

cos LOP 

Solve this triangle for OQP, QOP, OPQ. 

CPF = 180° - (OPQ + OPL) 

EOD = 90°-(QOP + LOP) 


Then 










CHAPTER VI. 


REVERSED CURVES. 


118. It is very undesirable that reversed curves should be 
used on main lines, or where trains are to be run at any con¬ 
siderable speed. The marked change in direction is objection¬ 
able, and an especial difficulty results from there being no 
opportunity to elevate the outer rail at the P.R.C. The use of 
reversed curves on lines of railroad is therefore very generally 
condemned by engineers. For yards and stations, reversed 
curves may often be used to advantage, also for street rail¬ 
ways, and perhaps for other purposes. 


119. Problem. Given the perpendicular distance between 

parallel tangents, and the common radius 
of the reversed curve. 

Required the central angle of each curve. 



Let AH and BD be the par¬ 
allel tangents. 

ACB the reversed curve. 

HB = p = perpendicular 
distance between tan¬ 
gents. 

Draw perpendicular NM. 

Let AOC = BPC = I r . 


Then vers AOC = AN = BM = iJ-IB 

AO PB AO 


120. Problem. 


r 2 P 

vers I r = A 
R 

Given p, I r . 
Required R. 


R = 


i P 


vers I r 

64 


(71) 


(72) 









Reversed Curves. 


65 


v" 

121. Problem. Given the perpendicular distance p between 
parallel tangents , the chord distance d between P.C. 
and P. T. ; and one radius Pi of a reversed curve. 
Required the second radius R 2 . 

Let ACB = reversed curve, 
e AH, DB parallel tangents. 

AB —d BH = p 
OA = Ri and PB = i? 2 
Connect AC and CB. 

AOC = BPC and ACO = PCB 
ACB is a straight line. 

Draw MP parallel to AB, OK 
perpendicular to MP. 

MP = AB and AM = BP 



OM : MK = AB : BH 

Ri -p _Z?2 \ d — d p 

When R 1 = R 2 = R 


or E l + R 2 = f 

2 p 

R = — 
4 p 


(72 A) 
(73) 


122. Problem. Given R and p. Required d. 


From (72 A) d = V'2 (R i + i? 2 ) P (73 A) 
When Ri = Ro = R d= V4 Rp = 2 V Rp (74) 


123. Problem. Given the perpendicular distance between 

two parallel tangents, and the central an¬ 
gle and radius of first curve of reversed 
curve. 

Required the radius of second curve. 



Let ACB = reversed curve 


HB = p ; AO = R i; PB = i? 2 
AOC = CPB = I r 
Draw perpendicular NCM. 

HB = AN + MB 


P 

R\ -f- R% 


= AO vers AOC -f BP vers BPC 
= Ri vers I r + i? 2 vers I r 

P 


vers I r 


( 75 ) 













66 


Railroad Curves and Earthwork. 


124. Problem. Given It i, i? 2 , P- 

Required 

from (75) vers ^ ^ (76) 

125. Problem. Given a P. C. upon one of two tangents not 

parallel, also the tangent distance from 
P.G. to V, also the angle of intersection, 
also the unequal radii of a reversed curve 
to connect the tangents. 

Required the central angles of the simple 
curves, and tangent distance, V to P. T. 



Let AV = T\ = given tangent distance 
ACB = required curve 


3 P 0 _ \ | required angles 

BV = T -2 = required tangent 
distance 


A = given P. C. 

V = vertex 
AVT = I 
AO = i?i 1 . 

PB = B, j glVen radn 
VT = second tangent 


Draw arc AL, also perpendiculars OL, AS, AK. 

Then LT = p = perpendicular distance between parallel tan¬ 
gents and by (75) p = (R i -f i? 2 ) vers LOC 

LT = LK + AS 


(Ri + Rf) vers LOC 
(i?i -f Rf) vers / 2 

vers I 2 

h 


AO vers AOL -f AV sin AVS 
Ri vers I + T x sin I 
R x vers / + T\ sin I 

T) 71 


(77) 


BV = VS + AK - TB 

T -2 = Ti cos I -f R x sin I — (R x -f i? 2 ) sin /> 


( 78 ) 










Reversed Curves. 


67 


126. Problem. Given BV instead of AV, and other data as 

above. 

liequired I x , / 2 , etc. 

Draw perpendiculars PH, BF, BG. 

UH = p = perpendicular distance between parallel tangents. 


UH = FH + GB 


(i?i + i? 2 ) vers R = R 2 vers I + T» sin / 

vers /, = gi vers/+ 7’ 2 sin/ 
111 4- Ri 


(70) 


Ti — Ti cos 1 4- H 2 sin / -f ( Hi 4- H 2 ) sin (80) 


Many problems in reversed curves can be simply and quickly 
solved by using the available data in a way to bring the problem 
into a shape where it becomes a case of parallel tangents with p 
known, and which can be solved by (75). 

This is true particularly of sidings and yard problems. 

127. Problem. Given the length of the common tangent and 


the angles of intersection with the sepa¬ 
rated tangents. 


Required the common radius of a reversed 
curve to join the two separated tangents. 


i 



Let V a Vb = common tangent 


P AVa, BV b = separated tangents 


ACB = required curve 
LV a C = I A ; MV e B = I B 


V A V B = l 

V A V B = V A C 4- V B C 


l = H tan \ I A + R tan \ I B 


M 


R = ---— r (81) 

t;i n 1 7. 4- t.n.n 1 T n 


tan \ ] A 4- tan | I B 


An approximate method is as follows : — 

Find l\i for a 1° curve ; also T B i (Table III.). 


n — ~F Tbi 
“ ' VaVb 


Then 


(approx.) 







68 


Railroad Curves and Earthwork. 


128. Problem. Given two points upon tangents not parallel , 
the length of line joining the two points, and the 
angles made by this line with each tangent. 

Required the common radius of a reversed curve 
to connect the two tangents at the given points. 



Let A and B be the given 
points. 

AL, BM = given tangents 
ACB = required curve 
LAB = A ; MBN =B 
AB = l 

AKO = PKB = K 
Draw PS perpendicu¬ 
lar and OS parallel to AB. 

Also OE perpendicu¬ 
lar to AB. 


Then 


PS = PU + SU 


OP sin POS = PB cos BPU -f AO siirOAB 
2 R sin K = R cos B + R cos A 

sin K = ° os A + cos B (82) 


AB = AE + EU + UB 

l = R sin A -f 2 R cos K + R sin B 

R =- l - - (82 A) 

sin A -f sin B + 2 cos K 

Values of sin A , sin B , and cos K may be taken out of table 
of nat. sines and cosines at the same time that cos A , cos B, and 
sin K are taken for (82). 






CHAPTER YU. 


PARABOLIC CURVES. 

129. Instead of circular arcs to join two tangents, parabolic 
arcs have been proposed and used, in order to do away with 
the sudden changes in direction which occur where a circular 
curve leaves or joins a tangent. Parabolic curves have, how¬ 
ever, failed to meet with favor for railroad curves for several 
reasons. 

1. Parabolic curves are less readily laid out by instrument 
than are circular curves. 

2. It is not easy to compute at any given point the radius of 
curvature for a parabolic curve ; it may be necessary to do this 
either for curving rails or for determining the proper elevation 
for the outer rail. 

3. The use of the “ Spiral,” or other “ Easement,” or “ Tran¬ 
sition ” curves secures the desired result in a more satisfactory 

x way. 

There are however many cases (in Landscape Gardening 
or elsewhere) where a parabolic curve may be useful either 
because it is more graceful or because, without instrument, it is 
more easily laid out, or for some other reason. 

It is seldom that parabolic curves would be laid out by 
instrument. 

130. Properties of the Parabola. 

(a) The locus of the middle points of a system of parallel 
chords of a parabola is a straight line parallel to the axis of the 
parabola (i.e. a diameter). 

(/;) The locus of the intersection of pairs of tangents is in 
the diameter. 

(c) The tangent to the parabola at the vertex of the diameter 
-is parallel to the chord bisected by this diameter. 

(d) Diameters are parallel to the axis. 

69 


70 


Railroad Carves and Earthwork. 


fe'l The equation of the parabola, the coordinates measured ' 
upon the diameter and the tangent at the end of the diameter is 


or 


y 2 = 4 p'x 


(83) 


131. Problem. Given two tangents to a parabola , also the 

position of P.C. and P . T. 
acquired to lay out the parabola by “ off- 
sets from the tangent 



Draw VG bisecting AB. 

Draw AX, BY, parallel to VG. 

Produce AV to Y. 

Then VG is a diameter of the parabola. 

AX parallel to VG is also a diameter. 

The equation of the parabola referred to AX and AY as axes is 


y°- = 4 p'x. 


Hence 


AV 2 : AY 2 = HV : BY 
AV 2 : (2 AV) 2 = HV : 2 GV 
1 ; 4 = HV : 2 GV 


Next bisect VB at D. 

Draw CD parallel to AX. 

BD 2 



BV 2 = CD . HV 


(84 


CD = 


HV 

4 


Then 




Similarly, make 
Then 


Parabolic Curves. 


71 


AN = NF 
HV 


FV 


KN = 


9 


EF = - HV 
9 


In a similar way, as many points as are needed may be 
found. 


132. Field-work. 

(а) Find G bisecting AB. 

(б) Find H bisecting GV. 

(c) Find points P, Q, and N, F, dividing AG, AV, proportion¬ 
ately ; also R and D, dividing GB and BV proportionately. 

Use simple ratios when possible (as |, etc.). 

(d) Lay off on PN, the calculated distance KN 

on QF lay off EF 
on RD lay off CD 

,/m HV 

In figure opposite, KN = — 


ef= 4 -hv 

Tor many purposes, or in many cases, it will give results 
sufficiently close, to proceed without establishing P, Q, R ; the 
directions o£ NK, EF, CD, being given approximately by eye. 
When the angle AVG is small (as in the figure), it will generally 
be necessary to find P, Q, R, and fix the directions in which 
to measure NK, EF, CD. When the angle AVG is large (greater 
than 60°) and the distances NK, EF, CD are not large, it wi 
often be unnecessary to do this. No fixed rule can be given 
as to when approximate methods shall be used. Experience 
educates the judgment so that each case is settled upon i s 

merits. 


Railroad Carves and Earthwork. 


9 


133. Problem. Given two tangents to a parabola , also the 

positions of P.C. and 1\ T. 

Required to lay out the parabola by “ mid¬ 
dle ordinates .” 



The ordinates are taken from the middle of the chord, and 
parallel to GV in all cases. 

Field-work. 

(«) Establish H as in last problem. 

(b) Lay off SE = \ HV ; also TC = \ HV. 

(c) Lay off UW = ATC, and continue thus until a sufficient 
number of points is obtained. 

The length of curve can be conveniently found only by meas¬ 
urement on the ground. 

Note the difference in method between § 85 and § 133. 

134. Vertical Curves. 

It is convenient and customary to fix the grade line upon the 
profile as a succession of straight lines ; also to mark the eleva¬ 
tion above datum plane of each point where a change of grade 
occurs ; also to mark the rates of grade in feet per station of 
100 feet. At each change of grade a vertical angle is formed. 
To avoid a sudden change of direction it is customary to intro¬ 
duce a vertical curve at every such point where the angle is 
large enough to warrant it. The curve commonly used for this 
purpose is the parabola. A circle and a parabola would sub¬ 
stantially coincide where used for vertical curves. The parabola 
effects the transition rather better theoretically than the circle, 
but its selection for the purpose is due principally to its greater 
simplicity of application. It is generally laid to extend an 
equal number of stations on each side of the vertex. 



Parabolic Carves. 


73 


135. Problem. Given the elevations at the vertex and at one 

station ( 100') each side of vertex. 

Beq Hired the elevation of the vertical curve 
opposite the vertex. 



Draw vertical lines AX, LGHV, MBY, and horizontal line ALM. 

Produce AV to Y. 

In the case of a vertical curve, the horizontal projections of 
AV and VB are equal, and here each equals 100 feet = AL = ML. 

Therefore AG = GB, and AV = VY 

VG is a diameter of the parabola. 

AX is also a diameter. 

vh=vg 

2 

Elev. H = 1 ^ j ^ L v _ A + Elev - v B + Kiev. V^ (85) 

This affords a simple and quick method of finding H when 
the vertical curve extends only one station each side of vertex, 
which is the most common case. Other methods or rules for 
vertical curves are used on various railroads, or by different 
engineers, and it will prove interesting and valuable to investi¬ 
gate such rules, to discover whether the resulting curve is a 
parabola, as will generally be found to be the case. When the 
vertical curve extends more than one station each side of the 
vertex, the following method is preferable, which is also appli¬ 
cable to the above case, and is in some respects preferable for 
that also. 






74 Railroad Curves and Earthwork. 

136. Problem. Given the rates of grade g of AV ; g' of VB ; 

the number of stations n, on each side of 
vertex , covered by the vertical curve; also 
the elevation of the point A. 

Bequired the elevation , at each 
station , of the parabola AB. 

Draw vertical lines ^ i 


x i 



Let 

«i = 

offset DD' at the 

first station from A. 


«2 = 

m 

m 

second “ “ A. 


«3 = 

etc. 


Then 

«2 = 

2 2 «i = 4 



«3 — 

3%i = 9 a\ 



a<2n : 

(2 ?i) 2 ax = 4 w 2 «i 

= YB 


YB = 

YC - BC 



4 = 

ng - ng’ 



= 

g - Q ' 

4 n 



Due regard must be given to the signs of g and g 1 in this 
formula, whether + or —. 
















Parabolic Curves. 


75 


From the elevation at A we may now find the required eleva¬ 
tions, since we have given g , 

and we also have oq 

a 2 = 4 


a 3 = 9 aq etc. 


A method better and more convenient for use is given below. 

DD" = g ; D'D" = 0 - oq 
EE" = 2#; E'E" = 2 0 — a 2 = 2 g — 4 eq 
VL = 3 g ; HL = 3 g - a 3 = 3 0 - 9 oq 

RT = 4 g ; TS = 4 0 — a 4 = 4 0 — 16 «i etc. 

Again, D'D" = g — oq = 0 — "i 

E'E" — D'D" = 2 g — 4«i-( g - flq)= 0-3a! 

HL — E'E" = 3 g — 9 oq -(2 0 - 4 oq) = 0 - 5 oq 

ST - H L = 4 £ - 16 «i - (3 £ - 9 «i) = $ - 7 «i etc. 

On a straight grade, the elevation of any station is found 
from the preceding, by adding a constant g. 

On a vertical curve, the elevation of each station is found 
from the preceding by adding, in a similar way, not a constant, 
but a varying increment, being for the 


1st station from A = g — cq 

2d “ “ A = 0 — 3 cq 

3d “ “ A = 0 — 5 cq 


changing by successive 
• differences of 2 aq in 
each case. 


The labor involved is not materially greater in many cases, 
for a vertical curve than for a straight grade. This method has 
the additional advantage that a correct final result at the end 
of the vertical curve makes a “check” upon all intermediate 
results. 



76 


Railroad Curves and Earthwork. 


137. Problem. Given g , ga\. Bequired n. 

From (86) n ~‘Tof’ ( 87 ) 

From practical considerations ci\ should not exceed 0.25 or 

’*74lfo& 0r n 7 g - g ' ' (88) 

Present practice indicates that a value of a\ as low as 0.05 is 
desirable ; in this case 

n-5(g~g') (88 A) 

With steep grades a value of a\ as high as 0.25 may fre¬ 
quently be necessary. 


138. Example. 

Given. Grades as follows : 


Sta. 

5 

10 

15 


Elev. 

117.00 
122.00 
124.00 


Rate 

+ 1.00 

+ 0.40 


Sta. 

5 

6 
7 


Then n — 5 (g — g') 
= 5 x 0.60 


= 3 


a i 


<J - (/’ 
4 n 


_ 0.60 
12 
- 0.05 


8 

9 

10 

11 

12 

13 

14 

15 


Elev. 



117.00 



+ 1.00 



118.00 



+ 1.00 

1.00 

= <7 

119.00 

- 0.05 

= «i 

+ 0.95 

0.95 

~g - «i 

119.95 

- 0.10 

— 2 a\ 

+ 0.85 

0.85 

e 

CO 

1 

0* 

120.80 

- 0.10 

— 2 a\ 

+ 0.75 

0.75 

g - 5 

121.55 

- 0.10 


+ 0.65 

0.65 


122.20 

- 0.10 


+ 0.55 

0.55 


122.75 

- 0.10 


+ 0.45 

0.45 


123.20 

End of curve 

+ 0.40 


= <J 

123.60 



4- 0.40 



124.00 




The elevation for Sta. 15 thus obtained agrees with the eleva¬ 
tion shown in the data. All the intermediate elevations are 
therefore “checked.” 























CHAPTER VIII. 


TURNOUTS. 

139. A Turnout is a track leading from a main or other 
track. 

Turnouts may be for several purposes. 

I. Branch Track (for line used as a Branch Road for gen¬ 
eral traffic). 

II. Siding (for passing trains at stations, storing cars, 

loading or unloading, and various pur¬ 
poses). 

III. Spur Track (for purposes other than general traffic, 

as to a quarry or warehouse). 

IV. Cross Over (for passing from one track to another, 

generally parallel). 

The essential parts of a turnout are 

1. The Switch. 2. The Frog. 3. The Guard Bail. 

1. Some device is necessary to cause a train to turn from the 
main track ; this is called the “ Switch .” 

2. Again, it is necessary that one rail of the turnout track 
should cross one rail of the main track ; and some device is 
necessary to allow the flange of the wheel to pass this crossing ; 
this device is called a tk Frog .” 

3. Finally, if the flange of the wheel were allowed to bear 
against the point of the frog, there is danger that the wheel 
might accidentally be turned to the wrong side of the frog 
point. Therefore a Guard Bail is set opposite to the frog, and 
this prevents the flange from bearing against the frog point. 

77 


78 


Railroad Curves and Earthwork. 


Frogs are of various forms and makes, but are mostly of this 
general shape, and the parts are named as follows: — 



M — mouth p = point 

0 = throat T = tongue 

WW = wings H = heel 


This shows the “ stiff ” frog. 

The “spring” frog is often used where the traffic on the 
main line is large, and on the turnout small. In the spring 
frog W'W' is movable. AD represents the main line, and W'W' 



is pushed aside by the wheels of a train passing over the turn¬ 
out. The “ Frog angle ” is the angle between the sides of the 
tongue of the frog = APB. 

Frogs are made of certain standard proportions, and are clas¬ 
sified by their number. 

The “ Number ” n of a frog is found by dividing the length 

of the tongue by the width of the heel • n - — = LH 

AB KD + AB 

140. Problem. Given n. Required Frog Angle F. 

A 

tan IF — = ^6 

PH 2 PH 

cot \ F = 2 n (PQ} 

























Turnouts. 


T9 


141. The form of switch commonly used at the present 
time is the “ split switch.” Fig. A shows the switch set for the 
turnout, and Fig. B for the main line. With the split switch the 




outer rail of the main line and the inner rail of the turnout 
curve are continuous. The switch rails, AB and CD, are each 
planed down at one end to a wedge point, so as to lie, for a por¬ 
tion of their length, close against the stock rail, and so guide the 
wheel in the direction intended. An angle, called the switch 
angle, is thus formed between the gauge lines of the stock rail 
and the switch rail, as DCE of Fig. B. The switch rails are con¬ 
nected by several tie rods, and one of the rods, near the point, 
is connected with another rod which goes to the switch stand S 
(or to a connection with the interlocking tower) from which the 
point of switch is thrown either for main track or for turnout 
as desired. The joint between the fixed end of the switch rail 
and the connecting rail, at B or D, is not bolted tight enough to 
prevent the slight motion of the switch rail necessary. The 
switch rail thus fastened at the end B is not spiked at all for its 
entire length, and acts as a hinged piece. Both rails thus move 
together, and through their entire length slide on flat steel 
plates provided for that purpose. The fixed (or hinged) end of 
this rail B is placed far enough from the stock rail to allow sat¬ 
isfactory spiking, frequently 5| or 6 inches. The length of 
switch rail has commonly been 15 feet, but it is not unusual to 
find a length of 19 feet or 24 feet or even more in some cases. 






























80 


Railroad Curves and Earthwork . 


142. Problem. Given , in a turnout , the gauge of track g , 

length of switch rail Z, the distance be¬ 
tween rails t, and frog angle F. 

Required the radius of turnout track and 
distance from movable end of rail to point 
of frog (or the lead ”). 



QF = QD+ DF 


Z +— 
tan IFD 


E- 


+ _ 9—i _ 

tan \ (F + 


( 01 ) 


Some prefer greater precision, and give weight to the fact 
that the frog is straight, not curved; it is not clear that this 
is worth while. 

Where the curve continues beyond the point of frog, the in¬ 
troduction of a short tangent (formed by the frog) hurts the 
alignment so far as appearance is concerned, and also makes 
less convenient the office work of laying out yards and other 









Turnouts . 


81 


turnout work. As a matter of fact, no advantage results so far 
as the easy running of the train is concerned. The path 
traveled by the center of a truck is here fixed, not by the line 
of the outer rail, as is the rule on curves, but by the position of 
the guard rail near the inside rail, and the actual path is likely 
to be a double-reversed curve of some sort, being fixed finally, 
in large part, by the gauge of the inside of the car wheels, and 
thus varying slightly with different wheels. 


143. Problem. Given, in a turnout, the gauge of track g , 

length of switch rail l, distance between 
rails t, length of frog, end to point, k, and 
the frog angle F. 

Required the radius of turnout curve , and 
the distance from movable end of rail to 
point of frog. 



Then following the methods of the 
preceding problem, 

sin S = - 
l 


R 



g — t — k sin F 
cos IS — cos F 


(92) 


E = l + 


g—t—k sin F 
tan %(F + /S’) 


-I- k cos 


F (93) 


It is the practice with some manufacturers of frogs and 
switches to bring the point of switch rail to substantially a 
razor edge, leaving it weak and liable to chipping off. Others 
plane the switch rail so that it has an appreciable thickness at 
its point E 5 the point, however, is not left blunt, but is shaped 
down through a short distance from the point so that the wheel 
will be safely carried by. The computation of the switch angle 
is changed. 

Let w = effective thickness of switch rail at the point. 

. u t — w 

Then sin S = —— • 

Formulas (90), (91), (92), (93) remain unchanged, but the 
lead E is the distance from the actual point of switch to the 
theoretical point of frog. 









82 


Railroad Curves and Eartlnvorlc. 


144. The variation in the values used for the length l of the 


switch rail, and of the offset t , makes it impracticable to tabu¬ 
late general values for the radius of the turnout curve. Each 
railroad has its standard turnout plans dependent upon the 
values of l , t , and F, and perhaps k. 

It has become the custom for the engineer of track to stake 
out the position of the frog point F. From this point F, a good 

track foreman will work back¬ 
ward and lay out the turnout ac¬ 
cording to the standard plan. 

For any continuance of turnout 
beyond the point of frog, a very 
common practice is as follows: 


B 



(a) Set the transit opposite the point of frog, at T. 

(/>) Lay off on the transit (on the proper side of 0°) the 
value of the frog angle F. 

(c) Sight in the direction TH, parallel to AB. 

(d) Turn off HTG = F. 

(e) The transit will then sight in direction TG with vernier 
at 0°. 

Any new curve desired may then be laid off conveniently by 
deflection angles, and this new curve will compound at T (op¬ 
posite F) with whatever curve extends backwards from the 
point of frog towards the point of switch. Where the line in 
advance of F is new location, TG is the basis for that location ; 
TG is either continued as £l straight line, or it becomes the tan¬ 
gent to a desired curve and the transit is already set on TG with 
the vernier at 0°. When the turnout is to connect with some 
definite outside line, especially a parallel siding, a warehouse 
track, or other line parallel to the main track, it is desirable and 
possible to set forth standard methods of solution and to de¬ 
termine certain formulas or processes which will apply. 


145. Two parallel straight tracks may be conveniently con¬ 
nected by a turnout in several important ways : 

I. By ( a ) the regular turnout curve to a point opposite F, 

(b) a tangent ending opposite the F of parallel track, 
and return by 

(c) the regular turnout. 




Turnouts . 


83 


II. 


in. 


By 


By 


(a) the regular turnout curve to a point opposite F, 

(ft) a tangent, and return by 

(c) a simple curve to parallel track. 

(a) the regular turnout curve to a point opposite F, 
and return by 

(b) a simple curve to parallel track. 


IV. 


By (a) the regular turnout curve which is carried beyond 
F, and reverse by 

( b ) the regular turnout curve to parallel track. 


V. 


By («) the regular turnout curve which is carried beyond 
F, and reverse by 

(b) a different turnout curve to parallel track. 


146. I. Problem. 



H 


Given the frog angle F, the gauge of 
track g, and the perpen¬ 
dicular distance between 
tangents p. 

Required the length l , 
of tangent between the 
two turnout curves. 

Let AB and CD repre¬ 
sent the rails on one of 
the parallel tracks and 
EG and HM of the other ; 


M 


AFNG will be one rail of the turnout or cross-over. 
Draw KTNL perpendicular to AB. 


Then TN = KL — KT - NL 

FN sin TFN = p — g — SN cos SNL 
l sin F—p - g - g cos F 
_ P - g - g cos F 
sin F 


(94) 


It is clear that the above solution holds good whatever be 
the turnout used, whether figured by § 142 or § 143 or otherwise. 

For a crossover between existing tracks, it will be convenient 
to calculate the distance from F to S. Both frog points can 
be located and the entire turnout staked out without transit. 

fp 

from (30) p S = l + (approx.) 









84 


Railroad Curves and Earthwork . 


147. II. Problem. Given g , p, jF. 



D G 


Bequired i? 2 0 / curve to 
connect parallel tangents. 

Let EFST and QCRD be 
the rails of the turnout; 
EN and QL, TW and DG, 
be the rails of the parallel 
tangents. 

Draw the perpendicu¬ 
lars US, SM. 


Then SU = LM = NT — NL - MT 


FS sin UFS = NT — NL — PS vers SPT 
l sin F = p — g — ^ | j vers F 

n (/ P — 9 — l sin F 

2 vers F 

p - g - ( 7?2- r, jvers F 

Similarly l = -A-iZ_ 

sin F 


(95) 


(90) 


148. III. Problem. Given F. n, p , g. 

, Bequired the radius of curve i? 2 , to 
connect the parallel tangents. 



If P. II. G. be taken at F. 
Then TPF = I r = F 


UT = US - TS 



PTvers TPF = US — TS 



vers 


F 


= p - g 


Pi 


g 

2 


p - g 

vers F 


(97) 


Second Solution. UT =p — q: PW = Bo 

2 

by (101) PW = UT 2 n* 

Ri — % — (p - g) 2 n 2 . 


(98) 


















Turnouts. 


85 


149 . For the proper solution of Case IV, when the first 
curve continues beyond the point of frog F before the P.B.C. 
is reached, it is desirable to develop the properties of a simple 
curved track whose outer rail is tangent both to the main track 
at C and to the frog at its point F. This curve will be called 
the “ Reference Curve.” 

Problem. Given gauge of track g , frog angle F. 

Bequired for the reference curve , B/ and DF = E. 

Let AT, CF, DR show the Reference Curve. 

Draw perpendicular FK; join CF. 


c 


K 


COF = F 

vers COF = ~~ 
F 0 


c 



vers F = 



DF = OF sin COF 



Second Solution. 


CFD = FCK = % F 
D F = CD cot CFD 
E — g cot % F 


— 2 ng 


( 100 ) 


FD 2 = FO 2 - DO 2 




E 2 = 2 B f x g 

E' 2 4 a 2 n 2 



B f —2 n 2 g 


( 101 ) 











86 


Railroad Curves and Earthwork. 


150. Problem. Given radius B f of Beference Curve , also 

radius B\ of some other curve tangent to 
the frog at its point F; also F, n , g. 
Bequired offset AH = a and distance Dl = d. 



I / 
Pi' 


Second Solution. 


Let OT = Bi and PC ±= B f + ^ 

Li 

A and C be points where FA and FC 
become parallel to DF. Then 

00 = ^ 7 ?/ + 

Al = 4- vers F 

AH = {Bf — Bf) vers F = a (102) 
Also 

Dl = DF - IF 

d = {B f - By) sin F (103) 


vers F 


By (101) 


Bf- 2 n 2 g ; 


whence CD = 
similarly AI = 


9 

OM 

27- 


Bf_ 

2 n 2 

^(approx.) 


A H= g 


El 

2 n* 


(approx.) 


By (100) DF = 2 ng 
Similarly 


IF = 


DF 

OM 

n 


Bf 

n 

B 

n 


= — (approx.) 


d — ——— (approx.) (104) 
11 / 

The approximation OM = Bi is ordinarily correct within y 1 ^ 
of 1 per cent. 

The distance from D to the point of frog as shown in (91) 
may be found thus : 

DF = 2 gn 


from (91) 


F= 1 + 


9 ~ t _ 

tan \ {F + S) 


DF — E — 2 gn — T? 4-—---1 CIOS'! 

L ' tan ^ {F + A> r ) J ^ 0j 

The point D can then be fixed with reference to the theoreti¬ 
cal point of the split switch as shown on standard turnout plans. 













Turnouts . 


87 


The practice of railroads in computing turnouts is not uni¬ 
form, nor upon any fully accepted lines. For the turnout 
proper there is some uniformity, and it is customary to have 
standard plans for the turnout from point of switch to point of 
frog, and to show on such plans the spacing of ties with their 
lengths, the lengths of the short rails used, and the position of 
the joints. Some also show offsets from the stock rail to the 
curved rail at points 10, 20, 80 feet, etc., from the point of frog. 
Such measurements result in much economy by saving the time 
of the track gang in laying out the work, and may result in a bet¬ 
ter track to run over. One railroad, in the case of slip switches, 
has all spike holes bored in the ties at the shop ; the tie spacing 
is also shown, so that the work goes in right and expeditiously. 
One road has the standard plan worked out and the frog located 
to make the lengths of short rails even feet, and thus avoid cut¬ 
ting rails ; the frogs are made with one leg shorter than the 
other, because the curved lead is slightly longer than the 
straight lead. The same result could be secured by setting one 
point of the switch slightly (2 ± inches) ahead of the other. 
Probably either device, while theoretically incorrect, will yield a 
turnout satisfactory to run over ; a consideration which is more 
important than any slight economy in cutting rails. The fact 
that the guard rail disturbs the smooth or uniform motion of the 
train along the curve suggests that extreme precision in figuring 
the turnout may not, after all, secure better practical results. 

Beyond the frog, uniformity of practice is lost. A common 
practice is that indicated in § 144, and this is convenient for 
field location. For general siding and yard work dealing 
mainly with parallel tracks, the method of p. 8(3 seems very de¬ 
sirable. While the treatment here is believed to be original in 
its details, it has its foundation in the practice of an important 
railroad, whose engineering is known to be among the best. 
Ordinary problems are solved almost as conveniently as they 
were upon the older stub switch basis. 

In laying out ladder tracks, as in §§ 153, 154, one chief 
engineer sets his transit at the point where LK produced (§ 154) 
intersects the main line rail (near F). The positions of the 
various frogs are staked out upon the line KL. The several 
distances required for this process can be readily figured, but 
no formulas will be given here. 


88 


Railroad Curves and Earthwork. 


151. IV. Problem. Given the common radius It of the 

turnout curves ; also F and p. 

Required angle AOS and distance FF'. 
Let AL = CM = a 
LH —p 
OB = R 

Find a by (102), p. 86. 
Then 

vers AOB = — a 



R 


(106) 


To find FF'; in triangle 
FOB, OB = R 

OF = i}+| 

FOB = AOB - F. 

Solve triangle for angle OFB and side F3 

FF' = 2 FB. 

F and F' are the theoretical points of frogs. 

152. V. Problem. Given the radii R\, i? 2 of two parts of 

the reversed curve ; also Fi, _F 2 , jt>, g. 

Required the angle FOS. 



Then by (76) 


T P — a\ — 02 
vers /.. = -- - -r 

Ri + R> 
FOS = I r - F x 
F'PS = I r - F 2 


AOF = Fi 
CPF' = F z 

OF = Ri + l 

PS = r 2 - 

LA = a\ 

CM = « 2 


(107) 

















Turnouts. 


89 


It is evident that instead of following the conditions of § 151 
or § 152 exactly by continuing the regular turnout curve beyond 
the point of frog, the problem may be made more general ; 
any reasonable radius may be used from F to S, although it 
would more commonly happen that either the regular turnout 
curve as found by (90), or the reference curve, p. 85, would be 
used. One radius may be selected for the first curve from op¬ 
posite the first frog point F to the P.P.O., and another from the 
P.R.C. to the second frog point F', provided only that the frog 
angle F> made by the second curve is one for which a frog can 
be readily furnished and that the P.R.C. shall come between 
the two frogs. 

Following the principle shown in § 120, p. 07, the solution is 
to be found by extending these curves (backward for Pi, for¬ 
ward for P 2 ) until they become parallel to the main tracks. By 
Pi and P 2 are meant here the radii OB and PB. 

The turnout is then arranged as follows: From the point of 
switch to the point of frog of each of the turnouts, the standard 
turnout (for the frog used) is laid. 

Beyond the frog point to the point of reversal it is possible to 
use Pi and P 2 or P /? or any combination of curves which may 
seem desirable. 

Whatever value of Pi (or P 2 or R f ) is used, it is necessary to 
consider this curve carried back (or forward as the case may be) 
until parallel to the main track, and find the yalue of a\ for this 
radius Pi (and of a 2 for P 2 ) and from these find 


vers I, 


p — (t\— a 2 

Pi -F P-2 


(107) 


When the reference curve is used beyond the point of frog, 
then «i and « 2 are each 0 and 

vers I r = ^T- (71) 

Rf 

The use of the reference curve as the basis for computations 
such as described probably gives the simplest results so far as 
computations are concerned. While ease of calculation should 
not control where convenience of construction or operation con¬ 
flicts, the method of computation shown here appears not to 
conflict with satisfactory construction or operation, and the 
same is true of the problem following. 



90 


Railroad Curves and Earthwork. 



also BC sin CBE = CE 


153. Problem. Given 
for tracks as shown in 
figure, the radius Bf of 
reference curve , also the 
perpe ndicular d istances 
between tracks p , p', p". 

Bequired BC, CD. 


BC sin AOB = p' 
BC = 


P* 


sin AOB ’ 


and CD = 


u 


_ P _ 

sin AOB 


(108) 


154. Problem. Given By, a, p, p', g , for standard turnout 

curve. 

Bequired AOK and KL. 


Let AK be the outer rail of standard turnout curve and By the 
radius of this curve, then 



— 2 By -}- a 


Then by (70) 
by (108) 


vers AOK — ———- (109) 

2 By + a 

i/i _ P r ' 

sin AOK 


The turnout with center at Q is also the standard turnout 


curve. 




















Turnouts. 


91 


155. Example. Given g = 4.708, p = 14, p ' = 13, n = 9, 

l = 15, t = 0.5, station of F = 91 4- 60. 

Required AOB = /,., BC, and station of 
P.C. of turnout curve. 

Split sic itch. § 151. 

sin S = ~ =0.03333 ; £-£=4.708-0.50=4.208 log 0.624076 
lo - 

S = 1° 55' 1° 55' cos = 0.99944 

Table XXII F = 6° 22' 6° 22' cos = 0.99383 



0.00561 log 7.748963 

£ (F + S) = 4° 08J' tan 8.859809 

750.1 log 2.875113 

g — t log 0.624076 

2 4-? 

58.1 log 1.764267 

2 

l = 15.0 

747.7 

to 

IT- 

II 


Reference Curve beyond F. § 153. 


R f — 4.708* x 2 x 9 

X 9 = 762.75 log 2.882382 

2 g= 9.416| 

\p — ~i log 0.845098 

9 

vers 7.962716 

Ef = 84.750 x 9 = 762.75 

4 = 7° 46' sin 9.130781 

4 = 7° 46' 

p' = 13 log 1.113943 

F= 6° 22' 

BC = 96.2 log 1.983162 

- ,F = 1° 24' 



The station of P.C. will be 91 -f- GO — 84.8 = 90 + 75.2. 
Turnout Curve beyond F. § 154. 



2 = 1495.4 




« = 0.1 

p = 14 

log 1.146128 

n xhx2 

1495.5 


log 3.174787 

9)747.7 


II 

0 

Ox 

i—* 

vers 7.971341 

9)83.08 = — 1 = 

E i 4 = 7° 51' 


sin 9.135387 

2)9.23 M 

F = 6° 22' 

/>' = 13 

log 1.113943 

4.61 
g ~ 4.71 
a = 0.10 

1 

II 

i— -* 

O 

tc 

BC = 95.2 

log 1.978556 


The station of P.C. will be 91 -f 60 — 83.1 = 90 -f 76.9. 



















92 


Railroid Curves and Earthwork. 


156. Problem. Given the B of either curve of a double 

turnout for a split switch , the equal frog 
angles F and F' , the distance DB from 
point to point of switch—d ; also n and g. 

Enquired, the angle C of crotch frog at C. 

Let ACF and BCF' be rails of equal turn¬ 
outs whose curves become par¬ 
allel to DF at N and L. 

Let OC = PC = B + | 

Continue arcs to N and L; 
join PO. 

j I / Draw perpendiculars AD, OM, PM, KL, PL. 

j ; / From (102) find a — SK = TN. 

j/ / Then MO = MK + KS + NO + NT - TS 

^ = 7?-f^-ftz-p2?-F + ft — 9 

MO = 2 B + 2a =2(7? + a) 

MP = KL = DB = d. 

In right triangle OMP find MOP and PO. 

i pr\ i po 

In isosceles triangle PCO, cos COP = - = —- (HO) 

0C 

2 COP = C. 2 

157. For turnouts from curves where the split switch is 
used, simple and exact solutions are not readily reached. Desir¬ 
able results may be secured by considering the effect of the 
curved track upon the “reference curve” in forming what 
may be called the “ modified reference curve,” a simple curve 
which comes tangent both to the curved main track and to the 
frog at its point F. This “modified reference curve” turnout, 
for which we may use the notation B/ and 7>/, then corresponds 
to the “reference curve” used with straight track. Very con¬ 
venient and simple relations will be discovered to exist between 
these curves and of a sort which will suggest simple modifica¬ 
tions of the split switch results already found, in order to adapt 
them to curved tracks. The following four pages will be de¬ 
voted to the establishment of these relations between the “ ref¬ 
erence curve ” and the “ modified reference curve.” 









Turnouts. 


93 


158. Problem. Given main 
track of radius B m ; F, g , and n. 

Bequired radius B/ of the 
modified reference curve for a 
turnout inside of main track; 
also E. 

Let EB be the outer rail of 
main track. 

EF the outer rail of turnout. 

Join EF. 

Let EOF = 0; PFO = F. 

In the triangle EOF, we have 



EO = B m + | 


FO = Bm - \ 

EFO - FEO = EFO — EFP = F 
EFO + FEO = 180° - 0 

Then tan |(EFO + FEO) : tan i(EFO-FEO) = E0 + F0 : E0 “ F0 

— 2 Brn • g 


cot \ O 


tan \ 0 


tan \ F 


: cot \ F 


= 9 


: 2 B r 


a . „ q2 n 

tan i° = 2 ir m mtiF= 'nr 


m 


tan \0 = % 


( 111 ) 


m 


Similarly, FPH = F -\- O 

Join HF, and in triangle HPF 


tan 0)=fp 


B/ = 


gn 


tan-|(F + O) 


chord HF = E = 2 f B m - ^ j sin \ O 


( 112 ) 

(113) 











94 


Railroad Curves and Earthwork. 


159. Approximate Formulas. 


Let R f , Df— Radius and a Degree of Reference Curve to cor¬ 
respond to the given value of F or n. 

E m , D,„ = Radius and Degree of main track. 

Rf, Df = Radius and Degree of “ Modified Reference 
Curve.” 


Then from (111) (112) R m = - 


ng 


tan b 0 


; B/- 


ng 


tan h (O + F) 


also 


(101) R/ = 2 vPg — ng x 2 n = 


ng 


tan 2 F 


50 


sin \ D m = — - 
sin J D/ — 


50 tan l 0 


Rin ng 

50 50 t an \ ( 0 


F) 


Bf 


ng 


. , 50 50 tan 4 F 

sin J D f = = --— 

R f ng 


sin l D, n : sin ) 2 D/ : sin \ D f = tan \ O : tan \ ( 0+ F): tan h F 

D m : D/ : D/ = O: O + /? : F (approx.) 

D m + Df : D/ = O + F: 0 + F (approx.) 

Df — Dm + Df (approx.) (114) 

Again, (113) HF = E = 2^R« - |)sin \ 0 

But f is small compared with R mi and may be neglected, and 
for small angles sin \ 0 — tan 0 (approx.) 

HF = E = 2 R ni tan h 0 (approx.) 

(Ill) J57 = 2 gn (approx.) (115) 

This agrees with (100) E = 2 ng (turnout from straight track) 


The above formula and (114), while approximate, are the 
formulas in general use. 

It is difficult in practical track work to secure results more 
precise than would be obtained by the use of these approximate 
formulas. 










Turnouts. 


95 


ISO. Problem. Given main track of radius i? TO , also F, g , n. 

Required radius JR/ of modified reference 
curve for a turnout outside of main track. 

I. When the center of turnout curve lies 
outside of main track. 

Let EB be the outer rail of main track, 
and H F the outer rail of turnout. 

Join HF. 

> 

Let HOF = O 

Then PFL = F 

In the triangle HOF 

FO=i^ + f 

HO = R m — | 

Also FHO + HFO = 180° - 0 

FHO — HFO = 180° - PHF — (180° — PHF — F) 

= F 

Then 

tan |(FHO + HFO): tan i(FHO - HFO)= FO + HO : FO - HO 
cot | O : tan \ F — 2 R m . g 

tan \ O = 

-Li in 

Similarly, OPF = F — O 
Join EF, and in triangle EPF 

tan l(F- 0) = jn 

-r. . & n 

Rf "tan*(F- 0 ) 
chord EF = E = 2^ + |)sin h 0 

Approximate Formulas. 

D/ = D f - Dm (approx.) 

E — 2 ng (approx.) 


(116) 

(117) 

(118) 



(HO) 




96 


Railroad Curves and Eartlnvork. 


161. II. When the center of turnout curve lies on the 
inside of main track. 

By a process entirely similar it may be shown that 

tan ; 0 = (120) 


it; = 


(jn 


/ tan ^ (0 — F) 
Jff=2^I? w + 0sin J 0 


( 121 ) 

( 122 ) 


Approximate Formulas. 

D f ' = D, n — D f (approx.) (122 A) 

E = 2 n<j (approx.) 


Example. Given a 3° curve on main line and a No. 9 frog. 

Required the degree of modified reference curve 
to the inside of the curve. 

Table XXII, Allen, shows for a No. 9 frog the 

degree of curve = 7° 31' = D/ 

The degree of main line = 3° 00' = D, n 

degree of turnout = 10° 31' = Df — D/ + D, n 

By precise formula 10° 32' = Df. 

The difference between approximate and precise results in 
the degree of curve is commonly small enough to have little 
effect in the short distance to the point of frog. 

In a similar way for a turnout on the outside of the main 
line, using a No. 9 frog, the degree of curve would be 

D f - D m = 4° 31'. 

In practice Df or Df would generally be taken as 7° 30' or 
4° 30' rather than 7° 31' or 4° 31'. 

162. The result that the proper degree of curve for the 
“modified reference curve” turnout is the degree of the ref¬ 
erence curve + or — the degree of main track is convenient, 
and is supported by § 70, where it was found that the “ tangent 
offset” is directly proportional (approx.) to the degree of 
curve, and that the offset between two curves of different d^- 





Turnouts . 


97 


gree is equal to the “ tangent offset ” for a curve whose degree 
is the difference between the degrees of the two curves. 

Applied to the turnout, and referring to the numerical ex¬ 
amples on pages 91 and 98, if the distance Ef = 84.8 (p. 91) 
and the outer rail of the 7° 31' reference curve (p. 96) deflects 
by g (the gauge of track) in passing from F. C. to F, it would 
follow from § 70, that in 84.8 ft. a 10° 31' curve would deflect 
from a 3° curve by the same amount g. 

Furthermore the central angle of the modified reference curve 
would be increased in 84.8 ft. by the same amount that the main 
track was curved in the same 84.8 ft. (approx.) and there would 
be no material change in the frog angle F in consequence. 

Another less mathematical, but very useful illustration is 
this: If we conceive the straight main track and the reference 
curve to be represented by a model where the rails are made of 
elastic material, then it will follow that if the main track rails 
be bent into a circular curve with the turnout inside, then the 
rails of the reference curve will be bent into a sharper curve, 
and, it is reasonable to believe, a curve sharper by the degree 
of curve D m into which the straight track is bent. That the 
modified reference curve is really sharper by just that amount 
(approx.) is clear by § 159 and § 70. 

Similarly when the straight track is bent in the opposite 
direction, then the modified reference curve will become flatter 
than the reference curve, and by the amount of D m . 

Applied to the split switch, when the turnout is from a 
curved main track, the proper procedure is to calculate all the 
data for a split switch turnout from a straight track, or lo use 
the data prepared for the standard turnout plan. The switch 
rail should then be bent to the degree of the main track D m , 
the curved rails of the turnout should be curved more sharply 
when the turnout is to the inside of the curved main track oi 
less sharply when to the outside. The lead distance from point 
of switch to point of frog should remain unchanged. The frog 
itself is short and the failure to curve it will mean very little. 
The switch rail may be long enough so that to secure a proper 
fit with the stock rail the switch rail should be curved. Some 
railroads specially bend the switch rails to fit the curves, but 
more commonly the switch rail is laid straight, and the train is 
relied on to bend it to a satisfactory fit. 


98 


Railroad Curves and Earthwork. 


163. Problem. Given the radial distance p between a given 

curved main track and a parallel siding , 
also frog angle F ( or number n), and 
gauge of track g. 

Required the radius Rf of second curve to 
connect point of frog with siding. 

I. When the siding is outside the main track. 

Let CM be the inner rail of 
the given main line. 

CFT inner rail of turnout. 

R m = radius of main line (cen¬ 
ter) . 

Rf = radius of turnout (cen¬ 
ter). 

p = TN = radial distance. 
Connect FT, FO. Join FS. 

Let FOT = O. 

In triangle FTO, FO = R m + | 

TO = R m - | + p 

also OFT + OTF = 180° - FOT = 180° - 0 

OFT - OTF — OFT — PFT = F 

Then 

tan $ (OFT + OTF) : tan $(OFT -OTF)= TO + FO : TO - FO 



cot \ 0 


tani0 = 2JJ. + p w ‘'* 


: tan J F 
cot A F = 


= 2 R m + p : p - g 


P - If ^ 1 ET _ P — 9 cot h F 


v 2 

r> _i_ 1 “ 

■L^m, ~T ~T 


tan J 0 = 


(p-u) n 
v 4 - hL 


Similarly 


FPT = F + 0 


In the triangle PFS, tan h (F + O) = —-; 




(123) 







Turnouts. 

% 

99 

p , P _ ( P-g)n 

2 tan ^ ( F -f O) 

(124) 

T f T 100 (Z 1 + O) 

Length of curve L =- - -- 

(125) 


Approximate Method. 

It might readily be shown that, if the entire turnout be cal¬ 
culated as if from a straight track, using the same values of 
n and p, and the degree of each curve (Zh, D>) be found; then 
it would be approximately true that, in the case of a curved 
main track, the degrees of the turnout curves required would 
be found by adding or subtracting D nl to or from Z>i and Do. 
The distances CF, FT would also be the same as in the turn¬ 
out from straight track. The demonstrations would follow in 
principle closely those given in reaching (114), (115). 


164. Example. 

Turnout from curve outside the main track. 
Let D m = 4 ; n = 9; p = 15; g = 4.7. 


Precise Method. 


tan 


0 = 


(p-g)n 10.3 92.7 

—V = 1440^ X9 = I440 
Bm + | 


v , P ( P~9) n 

12 2 “tan i(F+0) 

92.7 

— tan 6° 51' 45" 


92.7 

1440.2 

10 = 3° 40' 58" 
1^=3° 10' 47' 
%(F + O) = 0° 51' 45" 
92.7 
770.3 
\p= 7.5 

R 2 ' = 777.8 

D 2 ' = 7° 22' 17" 


log 1.967080 
log 3.158422 
tan 8.808658 

tan 9.080444 
log 1.967080 
log 2.880036 


L 


100(.F + O) 
D:' 


100 x 13°43'30" 


186.2 


7° 22' 17" 















100 


Railroad Curves and Earthwork. 


165. Approximate Method. 

In the case of a turnout from a straight main track, where 
n — 9 and p = 15, 

from (98) — i = (p — gr)2 n 2 

= (15.0-4.7)2 x 81 =1668.6 
i? 2 = 1676.1 ; Z >2 = 3°25'; F=6°22' (Table XXII.) 

L = —-= 186.3 for straight tracks 

3 2o' 

Do 1 = D-2 -f Dm 

= 3° 25'+ 4° = 7° 25' (7° 22' precise method) § 164 
L = 186.3 as with straight track (186.2 precise method). 



166. II. When the siding is inside 
the main track. 

In a similar fashion it may be 
shown, using this figure, that 


From triangle OFT 


tan i O — 


P 


(126) 


From triangle PFS 
Hz’ - U = 


(p - g)n 


2 tan | ( F — O) 


(127) 


- _ 100( F— O) 

1 / — -^;- 


ZV 


(128) 



167. III. When the siding is out¬ 
side the main track, but with the 
center of turnout curve inside of main 
track. 

Let EFS be the outer rail of main 
track. 


FT the inner rail of turnout. 











Turnouts. 


101 


From triangle OFT 

tan \0 = {p ~ g)n 

- *. + f 

(129) 

From triangle PFS 

-p i P (P-g)n 

2 2 tan ±(F+0) 

(130) 


100 (F+0) 

L ~ Bf 

(131) 


With both § 166 and § 167, approximate formulas may be 
used, the method being similar to that of § 163, p. 99. Experi¬ 
ence will determine when it will be sufficient to use the approxi¬ 
mate results, and when precise formulas should be used. 


168. Problem. Given the radial distance between a given 

curved main track and a parallel siding. 

The two tracks are to be connected 
by a cross-over, which shall be a re¬ 
versed curve of given unequal radii. 

Bequired the central angle of each 
curve of the reversed curve. 

Let AC center line of main track. 

AO = B m ; AP = Bf ; RQ = Bf 

Find a\ and a 2 by (102) 

Then in the triangle POQ 
find PO = B m + Bf -f a\ 

PQ = Bf + Bf 
OQ = OC + CB — BQ 
= Rm + P - B f - «2 

Solve for OPQ, PQO, POQ, then 
RQB. In practice this problem might 



take the following form : Given B 




p, g. Assume n (or F ) and if (or F'). From these calculate 
Bi and i? 2 , also Bf and Bf. Then solve as above. 







102 


Railroad Curves and Earthwork . 


Approximate Method. 

Where p is very small compared with B, n , the degree of curve 
used will frequently be found by the formulas (approx.) 

Di' — Dy — D m and Z> 2 ' = D> + D IU 

The length of each part may be found for a cross-over between 
parallel straight tracks, using the same values of n and p , and 
the same lengths used for this cross-over between curves. 

The process is similar in every way to that shown by example 
in § 164 and § 165. 

A similar method of treatment is applicable in all turnouts 
from curves if the distance between tracks is not too great. 

169. A form of switch formerly used is the “ Stub-Switch," 
which is formed by two rails, one on each side of the track, 
called the Switch Hails. One end of the rail for a short dis¬ 
tance (often about 5 feet) is securely spiked to the ties, the 
rest of the rail being free to slide on the ties, so that it may 
meet the fixed rails of either main track or turnout, as desired. 
These fixed rails, supported on a Head Block , are held by a 
casting, or piece of metal called the Head Chair , and upon 
which the switch rail slides. A Switch Bod connects the ends 
of the switch rails with the Switch Stand. One end of the rail 
is spiked down, so that when the free end is drawn over by the 
switch rod, the rail is sprung into a curve which may with 
slight error be considered a circular curve, tangent to the main 
line (if this be straight). The distance through which the free 
end of the rail is drawn or thrown by the switch rod is called 
the Throw of the switch. The free end of the rail is called the 
Toe , and opposite the P.C. of the curve the Heel of the switch. 

Knowing the throw t and the length l of the switch rail, we 
can deduce the radius B, or degree of curve Z>, and continuing 
this curve to the point of frog, we can readily deduce the angle 
between the rails or the Frog Angle necessary. 

It is more customary, however, having given the throw 
(5" — — C" are used on different roads), to assume either 

(1) the radius (or degree) of turnout curve, and from this 
find F (or n) and l ; or much more commonly 

(2) the number n (or angle F) of frog, and from this find B 
and then l. Table XXII. gives such data. 


Turnouts. 


103 


170. Problem. Given gauge of track g, frog angle F, and 
c h K throw of switch t. 

Required for a stub-switch , 
radius of turnout curve i?, 
length of switch rail l, and 
DF = F. 

Let CF, DQ be the rails of turnout. 

Draw KF, also HI at toe of switch. 

It is evident that the stub-switch curve is the 
same as the “reference curve” of § 149, p. 85. 
The same formulas will therefore apply. 



' ( 08 ^)iJ + f = 


2 vers F 


R = 2 tfg (101) 


E = f R -f 0sin F. F=2ng 


( 100 ) 


by (26) 
by (101) 


t = 


l 2 


(approx.) 


2 R 

l = V2 x 2 n-gt 


l = V2 Rt 
l = 2 n Vgt 


171. When there are two turnouts at the same point, one on 
each side of the main line, three frogs are necessary, the middle 
one being called the “ Crotch Frog .” This requires the use of 
at least two numbers of frog on any railroad. It is advisable 
that only two be used, and that all turnout curves be arranged 
to use one or the other of these two frogs. 




















CHAPTER IX. 


“Y” TRACKS AND CROSSINGS. 

172. In many cases where a branch leaves a main track, 
an additional track is laid connecting the two. This is called 
a “ Y” track, and the combination of tracks is called a “Y.” 


173. Problem. Given a straight main track HK, also the 

P. C. and radius of curve beyond the frog. 
Also radius R% of “ Y” track between the 
frogs. 

Required the distance HK from P.C. of 
turnout to P.C. o/“Y” track; also the 
central angles of turnout and of “Y” 
track to the point of junction. 



Let HK be the given 
straight main track. 

AB the turnout. 

CL the “ Y” track. 

Draw perpendicular NP. 


Let 


HK = NP = l 


AOB = I t 

CPL = I y = 180° - I t 

Find AH = a l ; KC = a z ; BL = c* 3 by (102) p. 86. 


Then cos AOB 

rrta T. — — KP _ i?i + a\ — Ro — # 2 

OB + BL + LP “ R x + R 2 + a 3 

l = (i?i + Iiz + « 3 ) sin It 


(132) 

(133) 








“ Y ” Tracks and Crossings. 


105 


174. Problem. Given a straight main track HEK, also the 

F.C., radius beyond the frog OB, and 
central angle AOB, of turnout curve con¬ 
necting with a second tangent BD ; also 
the radius PC of 11 F” track. 


acquired the distance HK from P.C. of 
turnout to P.C. of “F” track; also dis¬ 
tance BD from P.T. of turnout curve to 
P.T. o/“r” track. 



Draw parallel AV. Find a\ and « 2 by (102). 


Then BD = ED 

= ED 

= KP tan J CPL — 

= (P-2 + «2)tan \h — 

m = (i ?2 + # 2 ) cot J I\ — 
also HK = EK + 

l = (72 2 + $ 2 ) cot j 1\ + 


EB 

(VB + 
^AO tan J AOB -f 


EV) 

HA \ 
sin KEV/ 


( Bi tan £ Ii H-g^) 


EH 


^ 7?i tan \ h 



In case different frogs are used near D and K so that KC and 
DL are not equal, the same general method of solution will 
apply, but the demonstration and the formulas will be modified 


somewhat. 

Reference to § 185, p. 114, will show that 

EK = li 2 cot \ Iy H—;—— + - — 

sm 7i tan 7i 


ED = i ?2 cot l h + 


KC 


+ 


DL 


sin I\ tan I\ 













106 


Railroad Curves and Earthwork. 


175. Problem. In the accompanying sketch where 
HBC = main track. AD = turnout. 



LK =“7” track. 

Given HB = l; OB = B, n 
AP = Hi ; LQ = B 2 - 

Tiequired the points D and C. 

Find AH = a \; CK = a 2 

DL = di by (102) 

then PH = B\ 4 cii 
CQ — /?2 4 0,2 
DQ = i?2 4 ®3 


Considering 


Find 
Find also 
then 
then 


PH + BO as base 

HB as altitude 

OPH, then PO as hypotenuse 

PQ = B\ -f- _Z ? 2 4 QO = am 4 B 2 4 0,2 

POQ, OPQ, PQO 

BOC,APQ 


D and C will then be easily determined. 



176. In the figure where HBC is the main track and LK is 

the turnout, AD the “ Y ” track. 

Given OB = B, n ; KQ = B 2 
AP = Bi; BOC = O 

Bequired the points A and D. 

Find ai, a 2 , «3 by (102) 

Find QN, ON, then EP 
also EQP, EQ 
then EN = HB 
and PQO = EQP 4 OQN 

PQO determines position of L or D 
EPQ determines length AD and EN = HB fixes H or A 








“ 3 ^” Tracks and Crossings. 


107 


177. Problem. Given a curve crossing a tangent , 7?, g , < 7 ', 

angle C between tangent and curve. 

"Required frog angles at A, B, F, D. 

Draw AO, BO, CO, FO, DO ; also, MO perpendicular to CM. 

Then MO = R cos C 

9' 



MO - 


2 


n + l 


MO - 


r 

2 


M 

MO + ^ 


7? - 


(136) 


(137) 


(139) 


The rail length DF = (^R — ^ angle DOF; and BF = BL - FL. 


Example. Given C = 32° 28'; D = 8° g = 3 ; = 4' 8J". 

Required angle D and distance DF. 


i? 8 log=2.855385 
32° 28' cos = 9.926190 
MO=604.748 log = 2.781575 
\g<= 2.354 

607.102 log = 2.783261 
OF =715.28 log= 2.854476 
31° 55' 23" cos = 9.928785 

Table XX. 42'=0.0122173 
21' =0.00010118 
0.0123191 


i? 8 = 716.78 MO=604.748 

hg= 1.50 ig't= 2.354 
OD = OF = 715.28 602.394 

602.394 log = 2.779881 
715.28 log= 2.854476 
32° 37'44" cos 9.925405 
31° 55' 23" 

DOF= 0° 42' 21" 

2.854476 = log 715.28 = 7?- | 
log = 8.090579 

0.945055 log = 8.812 = DF. 





















108 


Railroad Curves and Earthwork . 


178. Problem. Given radii , It i, i? 2 , o/ two curves crossing 

at C ; angle at crossing C ; also g and g'. 

Required , /ro^r angles at A, B, D, E ; also 
lengths AB, BE, DE, AD. 



Having given OC = R i; 

OCP = C ; and PC = R 2 ; 
find in triangle OCP, the line OP. 

g 

2 

t_ 

2 

find in triangle OPA, angles 
APO, AOP, and OAP = A. 

Having givtyi OB = Ri + ^ 

A 


Having given OA = R\ 
also OP ; and PA = R 2 — 


also OP ; and PB = R 2 — 


ff' 


find in triangle OPB, angles 
BPO, BOP, and OBP = B. 


Then APB = BPO - APO, and AB 


=(*-fr) 


angle APB. 


The frog angles at D and E, and the lengths AD, DE, EB, 
may be calculated in similar fashion. 







CHAPTER X. 


SPIRAL EASEMENT CURVE. 


179. Upon tangent, track ought properly to be level across; 
upon circular curves, the outer rail should be elevated in accord¬ 
ance with the formula 

„ _ f /^ 2 
6 


in which e = elevation in feet 

g = gauge of track 
v = velocity in feet per second 
B = radius of curve in feet 
In passing around a curve, the centrifugal force 


c [Vv2 
32.2 B 


It is desirable for railroad trains that the centrifugal force 
should be neutralized by an equal and opposite force, and for 
this purpose, the outer rail of track is elevated above the inner. 
Any pair of wheels, therefore, rests upon an incline, and the 
weight W resting on this incline may be resolved into two com¬ 
ponents, one perpendicular to the incline, the other parallel to 
the incline, and towards the center of the curve. 

We 

The component p parallel to the incline will be p = — 

• 9 

It will be a very close approximation to assume that c acts 
parallel to the incline (instead of horizontally). The centrifugal 
force will be balanced (approx.) if we make 


p — c or 


We 


(J 


Wv 2 
32.2 R 


gv 2 

32.2 B 


w'hence 


e = 


(140) 








110 


Railroad Curves and Earthwork. 


In passing directly from tangent to circular curve, there is 
a point (at P. C.) where two requirements conflict; the track 
cannot be level across and at the same time have the outer 
rail elevated. It has been the custom to elevate the outer rail 
on the tangent for perhaps 100 feet back from the P.C. This 
is unsatisfactory. It is therefore becoming somewhat common 
to introduce a curve of varying radius, in order to allow the 
train to pass gradually from the tangent to the circular curve. 


180. The transition will be most satisfactorily accomplished 
when the elevation e increases uniformly with the distance l 
from the P.S. (point of spiral) where the spiral easement curve 

leaves the tangent; then - is a constant 

l 

or sra = A(a constant ) or m = w¥a 

Since g , v, A are constants, PI = C (a constant) 


Then 

also 

where 


Let 


Then 


PI = P c l c and P = 

t=-k or — = -t 
D D c D c l c 


(141) 
(141 .1) 


P c = radius of circle 

D c = degree of circular curve 

l c = total length of spiral. 

s = the “ Spiral Angle ” or total inclina¬ 
tion of curve to tangent at any point. 

s c = spiral angle where spiral joins circle. 

Pels = dl or ds = — 

P 


Ml 

Pci c 


2Pc? c 


( 142 ) 


Again 


dx — dl sin s 


and dy = dl cos s 







Cubic Spiral Easement Carve. Ill 

All values of s will generally be small, and we may assume 

sin s = s and 

then dx = sdl 

l 2 dl _ y 2 dy 
~ 2 Il c l c ~ 2 It c l c 

Integrating, x = (143) 

which is the equation of the “Cubic Parabola,” a curve fre¬ 
quently used as an easement curve. 

If, however, the approximation cos s = 1 be not used, the 
resulting curve will be more nearly correct than is the Cubic 
Parabola. In this case sin s = s 

dx = sdl = -^L 
2 RJc 

ji 

Integrating, x = —— (144) 

o ±\(Xq 

The resulting curve we may call, for the lack of a better 
name, the “Cubic Spiral” Easement Curve. 

181. The Cubic Parabola is well adapted to laying out curves 
by “ offsets from the tangent.” Modern railroad practice favors 
“deflection angles” as the method of work wherever practi¬ 
cable. In the case of an easement curve the longitudinal meas¬ 
urements are most conveniently made as chords along the curve, 

so that x = —represents a curve more convenient for use 
6 RJ C 

than is x = —^r- as wel1 as more nearl y correct. Evidently 

C 11,1c 

the properties of the two curves will be very similar. 

The proper easement curve is seen to be a curve of constantly 
(or at least frequently) changing radius. Searles, Holbrook, 
Talbot, and Crandall each have excellent curves of this char¬ 
acter. No important difference exists between them, except in 
the method and convenience of laying out the curve, and in this, 
the arrangement of tables is an important feature. Methods 
of use for the Cubic Spiral will be developed in the following 
pages. 


dy = dl 

y = l 








112 Railroad Curves and Earthwork . 


182. Given, in a Cubic Spiral, l, l c , B c 
Bequired s , s c , and u total deflection angles ” i, i c 



This is the expression for the central angle of the connect¬ 
ing circular curve whose length is one half the length of the 
spiral. It also follows that if the circular curve he produced 
back from C to K where it becomes parallel to AN, its length 

will be since KOC = CFN. Also AL = q = ^(approx.)(145 vl) 
Again for any point B on the spiral 

sin BAN = sin i = - (approx.) 

1 

i = - (approx.) = - * 


l 


% = 


Z 2 


6 BJJ 


But 

Whence 

Also 


s = 


C B c l c 
l 2 

2 BJc 


i = ~ and i, = — 


i : in = l 2 : U 


Sc 

3 


, l 

l = lc. — 


lc 

Also the back deflection ABG = BGN — BAN 

= s — i = Si — i = 2i 
ACF = 2 i 0 


(146) 


(146 A) 


Also 


(146 B) 










Cubic Spiral Easement Curve. 


113 


It will be observed that the Cubic Spiral has the following 
properties (some slightly approximate). 

(a) The degree of curve varies directly with the length from 
the P.8. 

(b ) The deflection angles vary as the squares of the lengths. 

(c) The offsets from the tangent vary as the cubes of the 
lengths. 

(d) The “spiral angle” at the point where the spiral joins 
the circular curve is equal to the central angle of a circular 
curve of the same degree and of a length one-half that of the 
spiral. 

(e) The deflection angle to any point on the spiral is one- 
third the spiral angle at that point. 


183. Given l , Z c , R c . Required y and y c . 
From (30) the excess of hypotenuse over base 

h 2 


e = c — a = 

2c 

Then in the Cubic Spiral, at any point on the spiral, let the 
excess de = dl — dy 

Z 4 dl 2 V dl 


from (30) 


integrating, 


de=^- = 


2 dl 2x4 R c H 2 dl 8 R 2 l c 2 
Z 5 


e=1—y— 
y-i- 
y c — ic 


40 Rc 2 lc 2 
Z 5 


40 RcH 2 

Z , 6 


40 R<Hc l 


= Z, 


U 


40 R 2 


(147) 
(147 A) 


184. Given R a , y c , s c . Required AL = q and LK = p. 

CN = ay and AN = y c 

AL = AN — OC sin COK or q = y c — R c sin s c (148) 

LK = CN — OC vers COK p = x c — R c vers s c (148 A) 

Values of i c from (146), i from (146.4), x c from (144), 
y c from (147^4), q from (148), p from (148.4) are computed 
and shown in Table XXXIII. of Allen’s Tables, as are also 
values of D computed from D c on the basis that D is directly 
proportional to the length of curve (141 A ). 









114 


Railroad Carves and Earthwork, 


185. Given /, Z, and B c or D c 

liequired the Tangent Distance T s . 

From Table XXXIII. 

(or by § 184) find q and p. P S. 

( a ) When the same spiral is used 
at both ends of the circular curve. 

AV = AL + LU + UV j 

= AL+ Kl + Ul tan UIV l 

— q T He tan J I 4 p tan h I 
T s — q 4 T c + p tan J I 

where T c is tangent distance for circular curve alone. 



(149) 


(&) When different spirals are used at the ends, separate 
values must be found 
for U V and for VW. 


H 


UV 

"TV 


H 

_T s' 


U 


Let Ul = pi 
IW = p 2 
HM = Pl 


\ 


w 


\ i X 

\\y \ 


In Fig. A VU = 


also 


HV - 
IW 

“sin VHI ~ 
T Sl = qi 4 de 4 

VW = VG - 
Ul 

— sin VGl 


isi — qi + T c — 


also 


l \ 

A. \ 

— ah 
\ 

\ 

\ 

HU 


Ul 

Pi 

tan UHI 

sin / 

Pi 

Pi 

sin I 

tan I 

WG 


IW 

Pi 

tan WGI sm 

Pi 

Pi 

sin I 

tan / 

Pi 

Pi 

tail / 

sin / 

- Pl 4 

Pi 

tan / 1 

sin I 

Pi 

Pi 


\ \ i 

' M 
v 


\ 


\ i\W 

—1- 

M I \ \ 

Fig. B. \ \ 


P 1 

tan I 


(149 A) 


Pi 

tan I 


(149 B) 


(149 A) 


Ts 2 = q-2 -f T e + 


sin I tan / 

Pi _ Pi 
sin / tan I 


(149 B ) 
















Cubic Spiral Easement Curve. 


115 


Example. Given a line as shown in sketch. 

Iiequired to connect the tangents by a 4° curve , 
with a spiral ISO feet long at each end. 



Find 


To Table III. 22° 14' 


Table IV. 

Table XXXIII. p = 0.94 
nat tan .j (22° 14') = 0.19649 
to nearest y 1 ^ ft. 0.19649 x 0.94 = 0.2 
Table XXXIII. q = 90.0 
i e = 1° 12' 

s c = Si c = S x 1° 12' = 3° 36' 

2 s c = 7° 12' 

I- 22° 14' 

I-2s c = 15 ° 02 ' 

4° ) 15.0333 

L = 


Deflection angles for spiral 
Table XXXIII. 
Transit at 43 + 01.0 
i = 0° 02' to 43 + 31.0 


0° 08' 
0° 18' 
0° 32' 
0° 50' 

t’c=l° 12' 


43 + 61.0 

43 + 91.0 

44 + 21.0 
44 + 51.0 
44 + 81.0 


T x = 1125.8 (4° 

281.45 
.05 corr. 

281.5 = T c 
90.0 = q 
0.2 = p tan £ I 
371.7 = T s 

V= 46 + 72.7 

3 + 71.7 = T s 
P.S. 43 + 01.0 

1 + 80,0 = l c 
P.C.C. 44 + 81.0 

3 + 75.8 = L 
P. C. C. 48 + 56.8 

1 + 80.0 = l c 
P.S. 50 + 36.8 


Deflection angles for circular curve 
Transit at 44 + 81.0 
back deflection to P. S. = 2 i c 


375.8 


for Ci = 19, 


I -2 So 



= 

2° 24' 

di 

= 0° 23' 

45 

2 

2° 23' 

46 


4° 23' 

47 


6° 23' 

48 


— 1° 08' 


’ 2 

7° 31' 

48 + 56.8 

02' 

= 7°31' 

Check 


2 


2 















116 Railroad Curves and Earthwork . 



from (145 A) Q = \ ( a PP rox -) 


CN = x c = 


1 3 
* c 


7 2 


6 ^ C Z C 6 _Z? C 

hV 


from (26) 
therefore 


CQ = 


2B 


for cubic spiral 


'2% = ikf ( a PP rox ^ for circle 


from (144) — = 


4 CO 

CN = g CQ = CQ + -- = CQ + QN (approx.) 

CQ = 3 QN = 3 KL = Sp (approx.) 

CN = 4 QN = 4 KL (approx.) 
x P 


X c lc S 

CN =2 3 TL = 8TL = 4KL 

TL = ~ = | (approx.) 


(149 C) 


(149 D) 


From CQ = 3 p the length of curve may be readily determined 
by method I or II, as follows : 

I. If the center of the circular curve KC be at 0, then 

KOC = CFN = 

, /r , r C Q 

vers KOC = or vers s c = ■ 

u K J-i>c 


100 s c 

~dT 


= L for circular curve KC 


%c 3 


from (146) 


l c = 2L 











Cubic Spiral Easement Curve. 


117 


Find other values of i by (146 A) i = 

The back deflection ACF = 2 i c 

If the chord lengths be taken = — the 

10 



required 


angles may 


be taken directly from Table VII. The actual numerical com¬ 
putation of angles proportional to the squares of the distances, 
while simple in principle, will be found somewhat burdensome 
in practice. The use of Table VII. is therefore recommended. 


I. Example. Given D c = 5° p = 0.90. 

Required length of spiral; also deflection 
angles. 


3p = 2.70 log = 0.431364 
_Z ?5 log = 3.059290 

s e = 3° 56' vers = 7.372074 

= i c = 1° 19' = 79' 

t) 

3° 56' = 3.933( 5° = D c 
78.67 = CK 

157.3 = L 


Table VII. opp. 79' 
gives values of i, 
point 1 1' 6 28' 

2 3' 7 38' 

3 7' 8 50' 

4 12' 9 1°01' 

5 20' 10 1° 19' 

Use 10 chords of 15.73 each. 


II. Find the chord length KC when CQ = 3p for the curve 
of given degree Z> c , by (26) or using Table VI. Take AL and 
length of curve KC equal to chord KC, and find correspond- 

ing central angle = s c . The deflection angle CAN = i c = ~ (146) 


IL Example. Given D c = 5 p — 0.90. 

Required length of spiral; also deflection 
angles. 

3 p = 2.70 = offset for 5° curve. 

2.70( 0.5 

5.4 corresponding offset for 10° curve. 

Table VI. shows chord corresponding = 78.7 = CK 

_5° = D c 

78.7 X 2 = 157.4 = l c s c = 3.935 = 3° 56' 

§ = i c = 1° 19' 

O 

Table VII. gives values of i as shown in Example above. 






118 


Railroad Curves and Earthwork. 


187. Problem. Given D c and l c . 

Bequired p and other data for spiral. 


From (23) 



3 p = B c vers KOC 


Other data will follow, as in Case I preceding. 

188. Field work of laying out Cubic Spiral , using Table 


XXXIII. 


(«) Select on ground (or fix) point L opposite the point K 
where the circular curve will 
become parallel to tangent. I 



( b ) Select from Table, length 
of spiral to join given circular 
curve. The tables will give, for 


the length, some multiple of - ^—• ~ L -^- 

30 ft. 

(c) Fix point A at P.S. of spiral (AL = <?), or set P.S. at 
proper distance T s from vertex V. 

(d) With transit at P. S. use chords of 30 ft. and deflection 
angles from Table, to set points on spiral, including the P. C. C. 
at C. 

(e) With transit at P. C. C. at C, turn vernier to O, and beyond 
O to measure angle FCA = 2 i c . 

(/) Sight on P. S. 

(g ) Turn vernier to O, and thus bring line of sight on tangent 
to curves at P. C. C. 

(h) Lay out circular curve as usual, or 

(i) If preferred, run out circular curve with transit at K, 
taking backsight on an offset from tangent = p. 

In original location, it may be sufficient to set stakes at A, at 
C, and opposite L, chaining between these points. For track¬ 
laying, points as close as 30 feet apart should be set. For 
making the cuts and fills, every other point may be sufficient. 





Cubic Spiral Easement Curve . 


119 


189. Given D c and p, 

Required to lay out Cubic Spiral by offsets from the tangent. 

Proceed as before and find x c — 4 p or use Table XXXIII. 
Find also l as in § 186. 

Pind other values of x at convenient intervals by formula 



This method now is less in use than formerly, since in general 
the spiral will be laid out by deflection angles in preference to 
offsets. 


Example. Given D c = 4°, p = 0.80. 

Required length of Cubic Spiral and offsets from 
the tangent. 


3p = 2.40 for 4° curve 
2.40(0 .4 

6.0 corresponding offset for 10° curve. 

Table VI. gives 83.0 = chord CK 
166.0 = l c 

l c = 166.0, x c = 3.20 =4 p = offset at end of spiral C, 

h = 83.0, x\ — 3.20 -T- 8 = 0.40 = offset at middle point L, 

h = 41.5, x 2 = 0.40 -4- 8 = 0.05 

h = 20.8, X\ = 0.05 h- 8 — 0.00625 

h = 62.2, x 3 = 0.00625 x 3® = 0.17 

h = 103.8, x 5 = 0.00625 x 5 3 = 0.78 

h = 124.5, x 6 = 0.00625 x 6 4 = 1.35 

Z 7 = 145.2, x 7 = 0.00625 x 7 3 = 2.14 


The “ cubic spiral ” will be laid out by measuring successive 
chords of 20.75 ft. each, and measuring the proper offset from 
the tangent. 


For the “ Cubic Parabola ,” 


the formula is 



whence 



The computations may be the same as for the cubic spiral. 
The successive distances of 20.75 will be laid off on the tangent 
and the offset laid off at right angles to the tangent. 




120 


Railroad Curves and Earthwork . 


190. It may occasionally (although not frequently) happen 
that the entire spiral cannot be laid out from the P.S ., and it 
will be necessary to determine deflection angles when the 
transit is at some intermediate point on the spiral. It will be 
desirable to occupy some regular chord point. 

In any Cubic Spiral, the degree of curve D increases uniformly 
with the length (141 H). Hence 
the degree of curve at I must be 
equal to the difference in degree 
between the circular curve and 
the spiral at 5 where length A 1 
= C5. 



Since the divergence in the degree of the spiral is the same 
for a given distance, whether this divergence be from the tan¬ 
gent AL or from the curve CK, it will naturally follow from the 
principles established in § 70, that the offset to the spiral for a 
given distance from C will be the same as the offset for the same 
distance from A, since the change in degree at corresponding 
points is always the same whether from tangent or curve. 

The same conclusion will be reached by referring to § 162 
near the middle of p. 97, where the elastic model and the “bend¬ 
ing process” is referred to; this bending process being there 
found to be correct (approx.) from the demonstration § 159, 
p. 94. If this principle be correct, it will follow that KT = TL, 
which may be considered an extreme case. That KT = TL is 
demonstrated (in 149 D ) to be correct is an additional assurance 

' T 

of the correctness of the principle stated above. 

It will further follow if E I and D 5 are equal, and at equal 
distances from A and C respectively, that the angles E A I and 
D C 5 will be equal (closely). For the offset divided by the 
distance gives approximately the sine of the angle, and since 
the angles are small the angles are proportional to the sines and 
so are in both cases equal. 

In other words, the divergence of any given spiral for a given 
distance, is the same either in offset or in angle, whether the 
divergence be from the tangent or from the circular curve. 




Cubic Spiral Easement Curve . 


121 


191. It will therefore follow that if at / 

\ / > 

any point B on the spiral ABC, the transit N.C/ / 7 J 

be set up and the line of sight be brought 7/ 


1 

A 

on the auxiliary tangent BG at that point, then the deflection 
angle to any forward point on the spiral will be the sum of 
(1) the “total deflection angle,” for the distance from B to 
that point, due to the circular curve HBJ, whose degree is the 
degree of the spiral at B ; and (2) the “ total deflection angle ” 
from the original tangent for that spiral for the same distance 
reckoned from the P.8. For any back point, the deflection 
angle from this auxiliary tangent will be the difference between 
these angles. 

The proper use of these deflection angles will allow the line 
of sight to be brought on the auxiliary tangent, as well as give 
means for setting all points on the spiral. Table XXXIII. gives 
values of D at chord points 30 ft. apart. 

Example. Required forward deflection angles from point 6 
on a spiral 270 feet long, to join 4° curve. 

Table XXXIII. gives D at point 6 =2° 40'. 

Deflection angle for 30 ft. on 2° 40' curve = 24'. 

The total angles will be at point 7, 24' + 01' = 25', 

8, 48' + 05' = 53', 

0, 72' + 12' = 1°24'. 

The tangent BG is found by laying off from chord AB, twice 
the forward deflection to point 6, or 2 x 48' = 1° 36'. 

The back deflections will be at point 5, 24' - 01' = 23', 

4, 48' - 05' = 43', 

3, 72' - 12' = 1° 00', 

2, 96' - 21' = 1° 15', 

1, 120' -33' = 1° 27', 

0 , 144' -48' = 1° 36'. 

The back deflection from point, 6 to P.8, also = 0° 48'x 2 = 1°36'. 




122 


Railroad Curves and Eartlnvork. 


192. Compound Curves. In the case of Compound Curves, 
it is proper and desirable that easement curves should be intro¬ 
duced between the two circular curves forming the compound 
curve. 



Problem. Given in a Compound Curve , D t , Z) s , p , or l. 

Required the Deflection Angles for a Cubic Spiral 
to connect the circular curves. 

( a ) Find by Table XXXIII. or § 186 the Deflection Angles 
proper for a Cubic Spiral to connect a tangent with a circular 
curve of degree = Di — D s . 

Let these = i’i, f 2 , 4, etc. 

( b ) Find the deflection angles to corresponding points on 
one of the circular curves, the auxiliary tangent for these being 
at the point where the Cubic Spiral leaves this circular curve 
(where the transit will be set). 

T . .i d\ d 2 

Let these = —, —, —, etc. 

A A A 

(c) The required total deflections from A will be for 

point 1 ^ + h point 2 ^ + i 2 

point 3 ~ -f 13 etc. 

A 

The required total deflections from C will be for 

• , d' i . . cV o 

point 1 ——point 2 ~ — i 2 etc. 

Ai A 

Similar procedure may be followed if it be desired to lay out 
the spiral by offsets. Convenient points may be set on the 
circular curves and the. offsets taken from either curve. 


Cubic Spiral Uaseinent Curve. 123 

193. Example. Given D t = 4°, D s = 7°, l c - 150. 

From Table XXXIII. use spiral 150 ft. long for Z> = 7°-4 0 = 3°. 
On 4° curve, deflection angle for 30 ft. = 30 x 0.3 x 4 = 0° 36', 
4° curve deflection + spiral deflection 

for point 1, 0° 36' + 0° 02' = 0° 38' 

36' 

2, 1° 12'+ 0°07'= 1°19' 

36' 

3, 1° 48' + 0° 16' = 2° 04' 

4, 2° 24' + 0° 29' = 2° 53' 

5, 3° 00' + 0° 45' = 3° 45' 

0° 36' x 5 = 3° 00' check. 

These are “total deflections” to the spiral when the transit 
is on 4° curve. 

When the transit is on 7° curve. 

On 7° curve deflection angle for 30 ft. = 30 x 0.3 x 7 = 1° 03', 
7° curve deflection — spiral deflection 

for point 1, 1° 03'- 0° 02'= 1°01' 

1° 03' 

2, 2° 06' — 0° 07' = 1° 59' 

3, 3° 09' — 0° 16' = 2° 53' 

4, 4° 12' - 0° 29' = 3° 43' 

5, 5° 15'- 0° 45'=4° 30' 

1° 03' x 5 = 5° 15' check. 


Field work. 

(а) Fix L or K in ground from topography or other practical 
requirements. 

(б) Assume l c and compute p ; or assume p and compute l c . 

(c) Fix A and C to be correct points on curve at distances 
l c from L or K. 

(d) Set transit at A. 

(e) Bring line of sight on auxiliary tangent at A. 

(/) Set off “total deflection” angles to spiral and run in 
spiral. 

(q) Note “ check ” on point C. 





124 


Railroad Curves and Earthwork . 


p 


194. Problem. Given I and B c for circular curve GHE, also 
and corre¬ 


sponding qfor 
a spiral to Jit P.S. 
the given curve. 

Required the distance 
BH = li through which 
the circular curve GHE 
must be moved in along 
VO to allow the use of 
this spiral ; also the 
distance GA — d from 
P.C. to P.S. 



BH = PO = KG 
h 
GA 


KL 


cos LKG 
P 

cos h I 


d- 


AL + LG 

AL + LK tan LKG 
q -f p tan \ I 


(150) 


(150 A) 


Problem. Given l, R c and h. 

Required p and d. 

p — h cos \ I 

q is found by Table XXXIII. or by § 186. 

d = q + p tan \ I. 

In re-running old lilies to introduce spirals, where an original 
circular curve is to be replaced by a spiral and a circular curve 
of the same degree, it is clear that the circular curve must 
necessarily be set in towards the center from H by a certain 
amount h. Practical considerations may often fix the distance 
h by which the curve must be moved. Tables VI. and VII. 
will be found of considerable value in revisions of line since 
they allow great flexibility in the selection of spirals. 








Cubic Spiral Easement Curve. 


125 


195. It may sometimes seem more desirable to change the 
radius of the circular curve so as to keep the new alignment in 
such position as to show as little deviation as possible from the 
old alignment and at the same time keep the length of line as 
nearly as possible unchanged. This may be accomplished as 
indicated in the figure below, where the line is carried outwards 
at B and inwards near D and L. 


Problem. Given I and B\ for circular curve DB ; also p 


of spiral; also BH =h 
measured along VO locat- A 
ing H through which new ~J7 
circular curve is to pass 
to allow the use of this 
spiral. 

Bequired the radius 
i? 2 = KP of the new curve 
KH; also q consistent 
with p and B 2 ; also dis¬ 
tance DA = d from P. C. 
to P.S. 


D L V 



PO = NO = OB + BH - PH 
— B i -f- h — _Z?2 
OM = DO - DM 

= DO - PK - KL 
= Bi - B 2 — p 
NM = NO - OM = h +p 


PO vers NOP = 
(B i - B 2 + h) vers \ I = 

B\ — B -2 -{* h — 


NM 

h -f p 

h + p 
vers \ I 


(151) 


Find q from p and B 2 , by § 186. 

DA = AL - DL 
= AL - MP 

d = q - (AT-1?2 + A)sin \I. (151 ^41 


Then 







126 


Railroad Curves and Earthwork. 


196. When it is necessary to keep the middle point H un¬ 
changed, on account of a bridge, or heavy embankment, or 
otherwise, it then becomes necessary to make part of the curve 
sharper, as CF in the figure below. The most practical method 
appears to be to assume the angle FOH of the original curve to 
remain unchanged and also the value of p and compute all 
other necessary data. 


Problem. Given 1 and R x of circular curve, also p of pro¬ 
posed spiral , also 
angle FOH = I x of 
the circular . curve 
which is to remain 
unchanged. 

Required the ra¬ 
dius i? 2 of new curve 
CF, to compound 
with original curve 
FH ; also q consist¬ 
ent with p and R 2 ; 
also the distance 
DA = d from F.C. 
to P.ti. 


OP vers NOP = NM = MD — ND 

= LP - KP =p 



(R i — Rf) vers Q /— If) = p 
R\ — R‘2 — 


P 


vers I- I x ) 


( 152 ) 


Find q from p and R 2 by § 186. 

Then DA = AL - DL 

- AL - MP 

d = q - (i?i - Rf) sin (i 1 - 7j) (152 A) 


"By making FOH = I x = 0, i? 2 becomes continuous from the 
fii-st spiral, through H and to its connection with the second 
spiral. 

Another practical method would be to assume R 2 and p and 
compute Ji, q, d . 







CHAPTER XI. 


SETTING STAKES FOR EARTHWORK. 

197. The first step in connection with Earthwork is staking 
out, or “ Setting Slope Stakes,” as it is commonly called. 

There are two important parts of the work of setting slope 
stakes : 

I. Setting the stakes. 

II. Keeping the notes. 

The data for setting the stakes are : 

(a) The ground with center stakes set at every station (some¬ 
times oftener). 

( b ) A record of bench marks, and of elevations and rates of 
grades established. 

(c) The base and side slopes of the cross-section for each 
class of material. 

In practice, notes of alignment, a full profile, and various 
convenient data are commonly given in addition to the above. 

198. I. Setting the Stakes. The work consists of: 

(«) Marking upon the back of the center stakes the “cut” 
or “ fill ” in feet and tenths, as 

C 2.3 or F 4.7. 

(b) Setting side stakes or slope stakes at each side of the 
center line at the point where the side slope intersects the sur¬ 
face of the ground, and marking upon the inner side of the 
stake the “ cut ” or “ fill ” at that point. 

1-7 


128 Railroad Carves and Earthwork. 


199. (a) The process of finding the cut or fill at the center 
stake is as follows : 

Given for any station the height of instrument = hi, and the 
elevation of grade = h g . 

Then the required rod reading for grade 

r g = hi — h g . (153) 

It is not necessary to figure h g for each station. 


Let h 9o = h g at Sta. 0 

h 9l = h g “ “ I 
hg 2 = h g “ “ 2, etc. 

Also use similar notation for r g . 

Let g = rate of grade (rise per station) 

Then h 9i = h ffo + g 

hg 2 — hg, g 
hg 3 = K + 9, et c- I 

r g Q = hi ~ hg o 
r g x = hi ~ hg x 

— hi — ( h 9o + g)= hi — hg o — g 
r g x = r g 0 ~9 ( 154 ) 

Similarly, r 9t = r 9i - g, etc. 


It will be necessary, or certainly desirable, to figure h g and 
r g anew for each new h t . It is well to figure h g and r g (as a 
check) for the last station before each turning point. 


Setting Stakes for Earthwork. 


129 


200. Example. h { = 106.25 

Sta. 0, grade elevation 100.00 
5, “ “ 105.00 

10, “ “ 107.50 

r g = 106.25 - 100.00 = 6.25 6.25 

9 X 

6.25 - 1.00/ = 5.25 


rate +1.00 
“ + 0.50 


r * = 

*v.= 


r H = 
% ~ 


5.25 - 1.00 

4.25 - 1.00 

3.25 - 1.00 

2.25 - 1.00 


Change in rate 


1.25 

0.75 


0.50 

0.50 


= 4.25 
= 3.25 
=z 2.25 
= 1.25 

= 0.75 
= 0.25 


It is found necessary to take a T.P. here, and we therefore 

find h a . — h 0r + 2 g 

1.00 = 106.00 


H + 
= 105.00 + 


r gi = hi - h 9l = 106.25 - 106.00 = 0.25 

Therefore all intermediate values r 9i , r 9t , etc., are “checked.” 


201. Having thus found r 9 , next, by holding the rod upon 
the surface of the ground at the center stake, the rod reading 

r c = LO is observed from 
,l i the instrument. The cut 
or fill 

c = OG = MN - LO 

= r, - r c (155) 

hg In the figure given the 
values of r g and c are posi¬ 
tive ; a positive value of c indicates a “cut,” a negative value 
of c indicates a “fill.” 

It can be shown that in the two cases of “ fill,” 

(1) When hi is greater than h g , and 

(2) When hi is less than h g , 

the formula given will hold good by paying due attention to the 
sign of r g , whether + or —. 








130 


Railroad Curves and Earthwork. 


202. ( b ) Setting the Stake for the Side Slope. 

(1) When the surface is level. 



M a G b N 


Let 


Then 


b 

c 


AB 

OG 


base of section 


center height 


BN 

EN 


AM 


s — ■— = — = side slope 


DM 

d = OD = OE = distance out 


d = GB+ BN 

= l b + s x DM = l b -f s x EN 
= 1 b + sc 


cm 


203. Setting the Stake for the Side Slope. 

(2) When the surface is not level. 

Here the process is less simple. 


ML N 



Let b = AB = base 

c = OG = center height (or cut) 


s = 


slope 







Setting Stakes for Earthwork . 

h r = EK = side height right 
hi = DH = “ “ left 

d r — GK = distance out right 
d, = GH= “ “ left 

Then d r = \ b + sh r 

. 

di= \b + slh ■ 

But h r and h t are not known. It is evident from the figure 
that h r > c and hi < c in the case indicated, and therefore 

d r >\b + sc 
di < \b + sc 


131 


(157) 


204. It would be possible .in many cases to take measure* 
ments such that the rate of slope of the lines OE and OD would 
be known, and the positions of E and D determined by calcula¬ 
tion from such data. But speed and results finally correct are 
the essentials in this work, and these are best secured by find¬ 
ing hi and li r and the corresponding d t and d r upon the ground 
by a series of approximations, as described below. 

Having determined c, use this as a basis, and make an estimate 
at once as to the probable value of h r at the point where the 
side slope will intersect the surface, and calculate d r = \ b -f sh, 
to correspond. 

Measure out this distance, set the rod at the point thus found, 
take the rod reading on the surface, and if the cut or fill thus 
found from the rod reading yields a value of d r equal to that 
actually measured out, the point is correct. Otherwise make 
a new and close approximation from the better data jnst ob¬ 
tained, always starting with h r and calculating d n and repeat 
the process until a point is reached where the cut or fill found 
from the rod reading yields a distance out equal to that taken 
on the ground. Then set the stake, and mark the cut or fill 
corresponding to h r upon the inner side, as previously stated 

Perform the same operation in a similar way to determine 
di = l b + shi, and mark this stake also upon the inner side with 
a cut or fill equal to hi. 



132 Railroad Curves and Eay'thwork. 

205 . It requires a certain amount of work in the field to ap¬ 
preciate the process here outlined, but which in practice is very- 
simple. It may impress some as being unscientific, and at first 
trial as slow, but with a little practice it is surprising how 
rapidly, almost by instinct, the proper point is reached, often 
within the required limits of precision at the first trial, while 
more than two trials will seldom be necessary, except in difficult 
country. 

206 . The instrumental work is the same in principle as at 
the center stake. 


ML N 



Let r T — NE = rod reading at slope stake right, 
then KN — NE = r g — r r = h r 

here r g is the same for center, right and left of section. 

In some cases it may be necessary to make one or more 
resettings of the level in order to reach the side stakes from 
the center stake. In this case, of course, a new r g must be 
calculated from the new h { . This introduces no new principle, 
but makes the work slower. 

207 . A “slope-board” or “level-board” may be used to 
advantage in many cases. In certain sections of country this 
might be considered almost indispensable. It consists simply 
of a long, straight-edge of wood (perhaps 15 ft. long) with a level 
mounted in the upper side. It is used with any self-reading 
rod. A rod quickly hand marked will serve the purpose well. 
Having given the cut or fill at the center, or at any point in the 
section, the leveling for the side stakes, and for any additional 
points, can readily, and with sufficient accuracy, be done by 
this “level-board,”, and the necessity for taking new turning 
points and resetting the level avoided. 



Setting Stakes for Earthwork. 


133 


208. II. Keeping the Notes. 

The form of note-book used for keeping the notes of slope 
stakes and of center cuts and fills, often called “ cross-section ” 
notes, is shown on the following two pages. 

The left-hand column for stations should read from bottom 
to top. 

The surface elevations in column 2 are not obtained directly 
from the levels , but result from adding to the grade eleva¬ 
tion at any station the cut or fill at that station, paying due 
attention to the signs. This column of surface elevations need 
not be entered up in the field, but may be filled in as office 
work more economically. 

The column of grade elevations consists of the grade eleva¬ 
tions as figured for each station. 

The figures marked + are cuts in feet and tenths, and those 
marked — are fills ; the figures above the cuts and fills are the 
distances out from the center, and the position in the notes, 
whether right or left of the center, corresponds to that on the 
ground. 

The columns on the right-hand page are used for entering, 
when computed, the “quantities,” or number of cubic yards, in 
each section of earthwork. 

209. The column “ General Notes ” is used for entering 
extra measurements (of ditches, etc.) not included in the 
regular cross-section notes; also notes of material “hauled” ; 
classification of material and various other matters naturally 
classed under the head of “ Remarks.” 

210. When the surface is irregular between the center and 
side stakes, additional rod readings and distances out are taken, 
and the results entered as shown for station 0 on p. 182, the 
section itself being as shown below in the sketch. 



Station 0 




134 


Railroad Curves and EarthivorJc 


211. Form of Cross-Section Book (left-hand page). 


(Date) 

(Names of Party) 

Base 20 ; 1 to 1 

14; 1£ to 1 



Surface 

Grade 





Station 




Cross-Section 



Elev. 

Elev. 





5 

97.1 

105.00 

18.4 

-7.9 


19.4 




-7.6 



-8.3 

+69.7 P. T. 

94.4 

104.70 

22.1 

-10.3 


23.0 




-10.1 



-10.7 

4 

96.9 

104.00 

19.3 

-7.1 


17.0 




-8.2 



-6.7 

+27.2 P.C. 

98.0 

103.27 

16.6 

-5.3 


12,4 




-6.4 



-3.6 

3 

98.1 

103.00 

16.0 

-4.9 


10,9 




-6.0 



— 2.6 

+87 

100.6 

102.87 

13.3 

-2.3 


7.0 




-4.2 



0.0 

+76 

102.S 

102.76 

10.3 

0.0 


11.9 




-2.2 



+ 1.9 

+64 

108.7 

102.64 

10,0 

+ 1.1 


13.2 




0.0 



+ 3.2 

+50 

106.4 

102.50 

13.4 

+ 3.9 


17.1 




+ 3.4 



+ 7.1 

2 

115.1 

102.00 

16.7 

+ 13.1 


26.7 




+ 6.7 



+ 16.7 

1 

117.7 

101.00 

22.7 

+16.7 

10.0 

22.2 




+ 12.7+17.2 

+ 13.1 

+12.2 

0 

109.2 

100.00 

18.0 

9.0 92 

8.5 18.4 

24.6 




*+• 8.0 4" 10.1 

+ 7.8 +14.7 

+ 14.6 





































Setting Stakes for Earthwork. 


135 


212. (Right-hand Page.) 


• 

Excavation 

Embank¬ 

ment 

General Notes 

L. Rock 

S. Rock 

Earth 




















136 Railroad Curves and Earthwork. 


213. Cross-sections are taken at every full station, at every 
P. C. or P. T. of curve, wherever grade cuts the surface, and in 
addition, at every break in the surface. In the figure below, 
showing a profile, sections should be taken at the following 
stations : —• 



At Stations 0, I, 2, 2 -f 52, 3, 4, 5, 5 + 80, 6, 

7, 8, 9, 9 + 29, 9 + 82, 10, II, 11+ 30, 12, 12 + 25 PC., 

13, 14, 15, 16, 17 P.P., 18. 

214 It is not necessary actually to drive stakes in all cases 
where a cross-section is taken and recorded, but in every case 
where they will aid materially in construction stakes should be 
set. It is best to err on the safe side, which is the liberal side. 

In passing from cut to fill, it is customary to take full cross- 

sections, not only at the point where the grade line cuts the 
surface at the center line of survey, but also where the grade 
cuts the surface at the outside of the base, both right and left , 
as in the figure below, which illustrates the notes on p. 134 ; 
full cross-sections are taken not only at stations 2 + 76, but 
also at 2 + 64 and 2 + 87. 


































Setting Stakes for Earthwork. 


137 


215. Stakes are actually set at the center G and at the 
point A, where the outside line of the base’ of Excavation cuts 
the surface, and at B, where the outside line of the base of Em¬ 
bankment cuts the surface. It is not customary to set stakes or 
record the notes for the points A' and B'. The stakes at A, G, and 
B are a sufficient guide for construction, and the solidities or 
“ quantities ” would in general be affected only slightly by the 
additional notes if they were made. When the line AGB crosses 
the center line nearly at right angles, it would not be necessary 
to take more than one section so far as the notes are concerned. 
It is well, however, to set the stakes A and B exactly in their 
proper position. 

216. Wherever an opening is to be left in an Embankment 
for a bridge or for any other structure, stakes should be set as 
in the figure below : — 



E 


At A and B (at the side of the base and top of the slopes AF 
and BH) stakes should be set marked u Bank to Grade ” ; and 
at F and H (at the foot of the slopes) stakes should be set 
marked “ Toe of Slope .” Where the bank is high, an addi¬ 
tional stake K at foot of slope may be set as an aid to construc¬ 
tion. The stakes at D and E should also be set as ordinary 
slope stakes. 

217. The “ level notes 11 proper, or the record of heights of 
instrument, bench marks, turning points, etc., used in setting 
slope stakes, are usually kept separate from the cross-section 
notes. One reason for this is that level notes run from top to 
bottom of page, while cross-section notes read from bottom to 
top of page. The level notes should be kept either in the back 














138 


Railroad Curves and Earthwork. 


of the cross-section book or in a level book carried for that 
purpose. Keeping these or any other notes on a slip of paper 
is bad practice. 

218. Earthwork can be most readily computed when the 
section is a “ Level Section that is when the surface is level 
across the section ; but this is seldom the case, and for purposes 
of final computation it is not often attempted to take measure¬ 
ments upon that basis. 

219. In general, in railroad work, the ground is sufficiently 
regular to allow of “ Three-Level Sections’ 1 '' being taken, one 
level (elevation) at the center and one at each slope stake, as 
shown by these notes, where Base is 20, and Slope \ to 1: — 

+ 2.0 + ' + 5.5 

The term “ Three-Level Section ” is usually applied only to 
regular sections where the widths of base in each side of the 
center are the same. In regular three-level sections the calcu¬ 
lation of quantities can be made quite simple. To facilitate 
the final estimation of quantities, it is best to use three-level 
sections as far as possible. 

220. In many cases where three-level sections are not 
sufficient, it may be possible to use “Five-Level Sections” 
consisting of a level at the center, one at each side where the 
base meets the side slope, and one at each side slope stake, as 
shown by the following notes : — 

Base 20, Slope 1 to 1, 

* 

22 - 7 10-0 , 16 n 10.0 22.2 

+ 12.7 +17.2 ' +13.1 +12.2 

The term “ Five-Lev el Section ” is usually applied only to 
regular sections where the base and the side slopes are the 
same on each side of the center. 

221. Where the ground is very rough, levels have to be 
taken wherever the ground requires, and the calculations must 
be made to suit the requirements of each special case, although 
certain systematic methods are generally applicable. Such 
sections are called “ Irregular Sections .” 








CHAPTER XII. 


METHODS OF COMPUTING EARTHWORK. 

222. In calculating the solidities or “quantities” of Earth¬ 
work, the principal methods used are as follows : — 

I. Averaging End Areas. 


II. Prismoidal Formula. 

III. Middle Areas. 

IY. Equivalent Level Sections. 
V. Mean Proportionals. 

VI. IIenck’s Method. 


223. I. Averaging End Areas. 

This is the simplest method : — 



Station 1 


Let Ao = area of cross-section at Station 0 


u 



l = length of section, Sta. 0 to St a. I 
$ = solidity of section of earthwork (Sta. 0 to I) 




140 


Railroad Curves and Earthwork. 


Then S= ^° ^ l (in cubic feet) (158) 

2 

* 

_ flo+yli . J_ cu ) 3 i c yards) (159) 
2 27 

As (158) is capable of expression 

S = A °\ +Ai \ 

it is practically based on the assumption that the solidity consists 
of two prisms, one of base Aq and one of base Ai, and each of 

a length, or altitude of -• 

2 


224. To use this method, we must find the area A of each 
cross-section ; the cross-section may be : — 

(a) Level. 

(b) Three-Level. 

(c) Five-Level. 

(cl) Irregular. 


225. (a) Level Cross-Section. 



a G B 


b = base = AB 

s = side slope = — = 

. AL BM 

c — center ht. = OG 


A = area of cross-section 







Methods of Computing Earthwork. 


141 


Then DL = EM = sc 

A = AB'x OG + DL x AL 

= be + sc 2 

= c(b 4- sc) (160) 

226. (b) Three-Level Section. First Method. 



Let b = base = AB 

s = side slope 
c = center ht. 
h r = side height EK 
ln= “ “ DH 

d r = distance out M E 
di = “ “ DL 

A — area of cross-section 

Then A — OGD + OGE + GBE + AGD 

= iOG X DL + lOG X ME + |GB x EK + JAG x DH 

=: \ c(di + d r ) + | - (hr + hi) 

c(di + dr) + (hi + h r ) 

& 


2 


( 161 ) 











Railroad Curves and Earthwork. 


142 


227. (6) Three Level Section. Second Method. 

M 


E 



Using the same notation. 
GB 


GV 

gv = gb = _a_ 

s 2 s 

OV = c -f GV = c + -- 

2s 

The triangle ABV is often called the “ Grade Triangle/' 
Area ABV = GV x GB 

4 s 

Area EODV = OV x — + OV x — 

2 2 


Let 


, 6 _\ di 4 - d r 

V 2s/ 2 


A = EODV — ABV 
_ , , b_\ ^L± d 

1 2 s/ 2 


D — di -f" d r 

A = (c + —- 
2s/ 2 


r_ ^ 

4 s 


62 
4 s 


( 162 ) 


In using this formula for a series of cross-sections of the 

same base and slope, — and — are constants, and the compu- 

2 s 4 s 

tation of A becomes simple and more rapid than the first method, 








Methods of Computing Earthwork . 


143 


228. (c) Five-Level Section. 

l o 



Use notation the same as before; in addition let 
f r — height M B 

J ,.= “ la 

Then A — AOB + DLOA + EMOB 

_ cb , fAr | J $1 

. _ cb -f- .1Ar 4~ fi ( h 


(163) 


229. (d) Irregular Section. 

The “Irregular Section,” as shown in the figure, may he 
divided into trapezoids hy vertical lines, as in Fig. 1; or into 
triangles by vertical and diagonal lines, as in Fig. 2. 




The triangles in Fig. 2 can be computed in groups of two, 
each pair having a common base (vertical). 

It will be seen that Fig. 1 requires 8 solutions and Fig. 2 only 
7 solutions of trapezoids or triangles. The computations can 























144 Railroad Curves and Earthwork. 


be made with substantially equal simplicity in either case, and 
after a little experience, directly from the notes without any 
necessity for a sketch. 

230. Another method which has been used for calculating 
irregular cross-sections is to plat them on cross-section paper, 
and get the area by “ Planimeter.' 1 ' 1 In very irregular cross- 
sections this method would prove economical as compared with 
direct computation by ordinary methods, but it is probable that 
in almost every case equal speed and equal precision can be 
obtained by the use of suitable tables or diagrams (to be ex¬ 
plained later) ; for this reason the use of the planimeter is not 
recommended. 

231. Having found the values of A for each cross-section, 
8 is found in each case by the formula above given, 

s _Ao±Ai m J_ p n cu ydg.) (169) 

It is found that this formula is only approximately correct. 
Its simplicity and substantial accuracy in the majority of cases 
render it so valuable that it has become the formula in most 
common use. It gives results, in general, larger than the true 
solidity. 

232. II. Prismoidal Formula. 

“ A prismoid is a solid having for its two ends any dissimilar 
parallel plane figures of the same number of sides, and all the 
sides of the solid plane figures also.” 

Any prismoid may be resolved into prisms, pyramids, and 
wedges, having as a common altitude the perpendicular dis¬ 
tance between the two parallel end planes. 

Let Ao and A\ = areas of end planes. 

M = area of middle section parallel to the end 
planes. 

I = length of prismoid, or perpendicular dis¬ 
tance between end planes. 

S = solidity of the prismoid. 

Then it may be shown that 

S^iAo + iM + A^l 

o 



Methods of Computing Earthwork. 145 


233. Let B = area of lower face, or base of a prism, wedge, 

or pyramid. 

6 = area of upper face. 

vi = middle area parallel to upper and lower faces. 
a = altitude of prism, wedge, or pyramid. 
s = solidity “ “ “ “ “ 


Then the area of the upper face b in terms of lower base B 
will be for 


Prism 
b = B 


Wedge 
6 = 0 


Pyramid 
6 = 0 


and the middle area m will be for 


Wedge 


B 


Pyramid 

B 

m = — 

4 


Prism 

m — B m = 

The solidity s will be for 
Prism 

, = aB = |.6J» = |(B + 4B + B) = |(B + 4m + 6) 

Wedge 

s = ^ = -- SB = -(b + — + oU-(5 + 4m + 6) 

2 6 6 V 2/6 

Pyramid 

s = ~ = a • 2B = -(B + — + 0^ = - (.B + 4 m + 6) 

3 6 6 \ 4/6 


Since a prismoid is composed of prisms, wedges, and pyramids, 
the same expression may apply to the prismoid, and this may 
be put in the general form 

S = (A* + tM+A{) 1 - 

6 

using the notation of the preceding page. 


(163 A) 




146 


Railroad Curves and Earthwork. 


234. A regular section of earthwork having for its surface a 
plane face is a prismoid. Most sections of earthwork have not 
their surface plane, and are not strictly prismoids, although 
they are so regarded by some writers. 

In this figure the lines E 0 Oo and EiOj are not parallel, and 
therefore the surface O 0 OiEiE 0 is not a plane. The most com¬ 
mon assumption as to this surface is that the lines O 0 O 1 and E 0 E 1 
are right lines, and that the surface O 0 O 1 E 1 E 0 is a warped sur¬ 
face, generated by a right line moving as a generatrix always 


E 



parallel to the plane 0 0 G 0 B 0 Eo and upon the lines O 0 Oi and 
E 0 Ei as directrices, as indicated in the figure. The surface thus 
generated is a warped surface called a “hyperbolic paraboloid.” 
It will be shown that the “prismoidal formula” applies also to 
this solid, which is not, however, properly a prismoid. 


235. In the following figure, which has perpendicular sides 
DoAoAiDi, EoBoBiF and the lines D 0 Eo and DiEi right lines, 


D, 



let b 0 = base = A 0 Bo 

by = “ = AjBi 

c 0 = center ht. = OoGo 
- DqAq ~h EqBq 
2 

Ci = center lit. = OiGi 
__ DiAi EiBi 
2 

l = length (altitude) 
of section = G 0 Gi 
Ao = area of D 0 A 0 BoEc 
A\ = area of D 1 A 1 R 1 E 1 
S = solidity 

























Methods of Computing Earthwork. 147 


Also use notation 5 Z , c z , A x for a section distant x from Gi. 
Then 

do = boCo = biCi 


x 


b x — bi 4- (bo — &i) — 

I 


c x — Ci (d Co) - — Ci 4* (c 0 — Ci) — 

l l 

A x = b x c x — j^5i 4 - (bo — bi) — J [ci -f- (co — Ci) 

& — ^1 + (Po — bi) [ci 4- (c 0 — Ci) 

= 6iCiZ + [5i(co-c 1 )+ci(f; 0 -fc 1 )]-^+ ^ a ~ 6l X c o" . c iy 

21 SI 2 

G & 1 C 1 4~ 3 5iCo 4~ 3 Mi + 2 b 0 C 0 
l — 3 ftiCi — 2 Z>ic 0 — 2 byCi 
6 — 3 t»iCi 

• . -f- 2 hiCi 

S= — (2 ^iCi 4- 2 boCo + 61 C 0 4" 5qCi) (164) 


x 


236. Apply the “ Prismoidal Formula 11 to tlie same section. 
The base and center height of the middle section are : — 


7, _ bo -f bi 

Urn .— ' ' 


Aq — boCo 


__ Co 4- Ci 
—--- 

2 

Ai = f^i'Ci 


M _ bo 4- b 1 x Co 4- Ci _ area 0 £ section 

2 2 

S= t -(Ao + i3f+A,) 

0 

~ ^ (paCo 4- boCo 4- frod 4- biCo 4- & 1 C 1 4- & 1 C 1 ) 
= - (2 b^i 4- 2 boCo + biCo 4- 5oCi) 


(165) 


This is the same as formula (164) found above to be correct 
for the warped surface. Therefore the “ Prismoidal Formula *’ 
(163) applies to the section shown in § 235. 









148 


Railroad Curves and Earthwork. 


237. The sections of earthwork commonly used in railroad 
work are bounded not by perpendicular sides, but by inclined 


planes. 


In the figure, suppose 
a plane to be passed 
through the line E 0 E 1 , 
cutting A 0 Bo at Po and 
A 1 B 1 at Pi. The pris- 
moidal formula applies 
to the solid E 0 P 0 B 0 B 1 E 1 P 1 
cut out by this plane, 
since this solid is a true 
prismoid. If the pris- 
moidal formula applies 
to the entire solid, and 
also to the part cut out, 
it must apply to the re- 



D 


maining solid D 0 A 0 P 0 E 0 E 1 P 1 A 1 D 1 , and this represents in form 
one side of a regular three-level section of earthwork in which 
D 0 Ao represents the center height and EoPo the slope. 

If the prismoidal formula applies to the section upon one side 
of the center, it applies also to the other side, and so to the 
entire section. 

238. The “ Prismoidal Formula ” is of wide application. 
Since it applies to prisms, wedges, pyramids, and to solids 
bounded by warped surfaces generated as described, it follows 
that it applies to any solid bounded by two parallel plane faces 
and defined by the surfaces generated by a right line moving 
upon the perimeters of these faces as directrices. It may also 
be stated here without demonstration that it also applies to the 
frusta of all solids generated by the revolution of a conic sec¬ 
tion as well as to the complete solids, for instance, the sphere. 

The prismoidal formula is generally accepted as correct for 
the computation of earthwork and similar solids, and the meas¬ 
urements of a section of earthwork are taken so as to repre¬ 
sent properly the surface of the ground if this be a warped 
surface of the sort described. The failure to use the prismoidal 
formula is explained often by the additional labor necessary 
for its use. 









Methods of Computing Earthwork. 149 

239. For “three-level” sections of earthwork, a result cor¬ 
rect by the prisinoidal formula may be secured, and the work 
simplified, by calculating the quantities first by the inexact 
method of “end areas,” and then applying a correction which 
we may call “ The Prismoidal Correction.” 

Let S e = solidity by end areas 

S p = “ “ prismoidal formula 

Then C = S e — Sp = prismoidal correction 
In the figure, § 235, 

S p = by formula (164) = - (2 biCy 4- 2 b 0 c 0 f b x Co + b 0 c x ) 

6 


S e — - (biCi + boCo ) ~ — (3 biCi + 3 boCo) 

2 0 



6 


6 


(5 1 - b 0 )(ci - c 0 ) 


Let D 0 A 0 = lid 


D1A1 = hi' 


E1B1 = hi 


Then 


C= l - (b l -b 0 )(’±± 


hi 4- hi' ho -Jr h 0 ' 


2 


) 


= f-(bi-b 0 ) (hi + hi' -h 0 - ho') 



When the solid assumes a trian- 
1 1 gular cross-section, as in the figure, 

'b, . 


ho' — 0 hi' — 0 


C = -(bi-b 0 )(hi-ho) (166) 






150 


Railroad Curves and Earthwork. 


240. If any solid be divided into a number ot solids each of 
triangular cross-section, the above correction may be applied to 
each such triangular solid, and the sum of the. corrections will 
be the correction for the entire solid. 



Let this figure represent a section of earthwork divided into 
three parts, as indicated by the lines D 0 G 0 , E 0 Go, D 1 G 1 , E 1 G 1 . 

Then, for the solid OoDoGoE 0 EiGiDiOi, 

° = 12 C ( Cl “ C °^ ^ ~ ^ + ^ Cl “ C °) 3 

= — (t‘i — Cq) (d ti + d ri — di Q — d ro ) 

Let I)i = di t + d rL and D 0 = d lo + d T() 

C "l2 Vi-<’o)(Z>i-A>) 


For the solid G 0 B 0 E 0 EiBiGi, 


(166) 



Similarly for the solid A 0 G 0 D 0 DiGiAi. 

Hence for the entire solid AoBoEoOoDoDiOiEiBiAi. 


( 167 ) 





Methods of Computing Earthwork . 151 


When l = 100 


Cloo = 


Since 


= 77 , ( C 1 ~ Co) (A — Do) in CU. yds. (168) 

C — S e — 

S p = S e - C (169) 


When (ci — c 0 ) (Di — Z>o) is positive , the correction (7 is to 
be subtracted from S e . 

When (ci — Co) (D\ — D 0 ) is negative , the arithmetical value 
of C is to be added to S e . The latter case seldom occurs in 
practice, except where C is very small, perhaps small enough 
to be neglected. 

For a section of length l, 


Ci = 




100 

l 


C: 


100 


= 7 ^ 0 $ 


100 


«ioo 


^ioo) 


(169 A) 


241. In general, for sections of earthwork, the prismoidal 
correction as given above applies only when the width of base 
is the same at both ends of the section. There are certain 
special cases, however, which often occur, and which allow of 
the convenient use of this formula for prismoidal correction. 
Referring to the figure on p. 186, and the corresponding notes 
on p. 134, the correction can be correctly applied in the case of 
the excavation from Sta. 2 + 64 to 2 + 76 as follows : — 
Compute S e , and then apply C , using at 

Sta. 2 + 64 D 0 = 28.2 

and at Sta. 2 + 76 Zfi = 11.9 = d ft 

or the distance out on one side only. This may readily be 
demonstrated to be proper if the correction to the right of the 
center be taken, using formula (167), and the correction to the 
left using formula (166), and the two corrections (right ami 
left) be added. 






152 Railroad Carves and Earthwork. 


242. Formula (106) can also be used to find the correction 
for the triangular pyramids (for excavation Sta. 2 + 76 to 2+87. 
and embankment 2 + 64 to 2 + 76), each end of the pyramid 
being considered to have a triangular section. A much simpler 
way to find the correction for a pyramid is this, 

G=S.-S, = \S. 

o 

as may readily be shown to be true for any pyramid, since 

S.= A l - 
2 

'%= A 

C = S.-S, = A l - = & (170) 

243. In the case of regular “ Five-Level Sections,” as shown 
in the figure, p. 143, the prisinoidal correction may be com¬ 
puted for each of the triangular masses bounded by 

1. AOB 2. OBE 3. OAD 

In the case of AOB, the prisinoidal correction will evidently 
be = 0, since Do — 0 = Zfi, and therefore Do — 7+ = 0. 

The correction for the mass bounded on one end by 

0 B E = c=T(/,.- (<*_„ -<?„,) 

and by OAD = C = ^ (/, o - /,_) (* 0 - d, t ) 

OBE and OAD differ but little from regular sections of earthwork 
in which b = 0. 

244. In the case of “Irregular Sections,” the prismoidal 
correction cannot with convenience be accurately employed. 
There are, however, several methods by which we may calcu¬ 
late a “prismoidal correction” which will be approximately 
correct. 

For the purpose only of calculating the correction, either of 
the following methods may be employed : — 


Methods of Computing Earthwork. 153 

1. Neglect all intermediate heights, and figure correction 
from center and side heights. 

2. Find level sections of equal area in each case, and figure 
correction, using the center heights and side distances of these 
level sections. 

3. Having c and D of the irregular section, either 

(а) retain c and calculate D , or 

(б) “ 1) “ “ c 

for a “ regular three-level” section of equal area, and use these 
values to calculate the correction. 

4. Plat the cross-section on cross-section paper, and equalize 
by a line or lines drawn in the most advantageous direction, 
and from the c and D thus found compute the correction. 

245. In relation to these methods : — 

No. 1 is most rapid and least accurate. 

“ 2 is less exact than 3 in most cases, and probably no more 
rapid. 

“ 3 is recommended as nearly equal to 4 in accuracy, and 
far more rapid. 

“ 4 would yield the most accurate results. 

The value of these approximate methods cannot be properly 
appreciated until certain rapid methods of computation are 
understood, as will be appreciated later. 

The results obtained by the methods shown above are ap¬ 
proximate only, but in most cases the resulting error would be 
small, or a small fraction only of the entire correction , which is 
itself generally small. 

246. The method of calculating by averaging end areas 
and applying the prismoidal correction will be found much 
more rapid than to calculate the middle area and apply the 
prismoidal formula directly. There is another advantage of 
importance in favor of the use of the prismoidal correction ; 
in a majority of cases for sections used, the method of “end 
areas,” is sufficiently accurate for all practical purposes, and 
from the use of the prismoidal correction the computer will 
soon learn to distinguish, by inspection merely, in what cases 
this correction need be applied. 


154 


Railroad Curves and Earthwork. 


247. III. Method of Middle Areas. 

This consists in calculating the area of the middle section 
(not the mean of the end areas), and assuming the solidity to 
be that of a prism having a base equal to this middle area, and 
an altitude equal to the length of the section of earthwork. 

Let M = middle area 

l = length of section 

Then S = Ml 

This method is not exact. It gives results generally less than 
the correct solidity. It is not sufficiently rapid to recommend it. 

248. IV. Method of Equivalent Level Sections. 

This consists in finding level end sections of equal area with 
the actual end sections from these calculating the level middle 
section, assuming the top surface connecting the level end sec¬ 
tions to be a plane; and then calculating the solidity of this 
prisinoid by the prismoidal formula. 

This method is not exact; it gives results less than the correct 
solidity. It is not sufficiently rapid to recommend it. 

249. V. Method of Mean Proportionals. 

This consists in assuming that the solid is the frustum of a 
pyramid, in which case all its sides would meet in one vertex. 
This method is not exact. It gives results always less than 
the correct solidity. 

250. VI. Henck’s Method. 

In connection with the prismoidal formula, it was stated that 
the most common assumption was that the upper surface is a 
warped surface of a certain kind which was there described. 
Henck's Method assumes otherwise ; that the upper surface is 
divided into plane surfaces by diagonals from the center height 
of one cross-section to the side height of the next, as shown 
in the figure, where the diagonals OiE (1 and O 0 Di are drawn. 


Methods of Computing Earthwork. 155 


Which way the diagonals shall be assumed to run is deter¬ 
mined on the ground by the shape of the surface in each case. 



The diagonals O 1 E 0 and O 0 D 1 divide the surface into four plane 
surfaces, D 0 O 0 Di, O 0 DiOi, O 0 E 0 Oi, E 0 OiEi- 


251. Let this figure represent the right-hand side of a section 



si = area G 0 B 0 BiGi x 


of earthwork, with the diagonal 
assumed to run from Eo to Oi. 
Join E 0 with Go, Bi and Gi. 
The entire solid may then be 
considered as composed of three 
pyramids having their vertices in 
a common point Eo- 

Using notation already famil¬ 
iar, the solidities of the three 
pyramids are as follows: — 

height at Eo 3 


s 2 = area G 0 O 0 OiGi x distance out to E 0 3 

_ Cq -f- Ci i 

2 3 

$ 3 = area O 1 G 1 B 1 E 1 x length of section -*■ 3 


1/6 













156 


Railroad Curves and Earthwork . 


Let S r = solidity of this right half of section 


S r — Si + s 2 4* s 3 


= l bl,l r« + l dr 0 lC ° + ^ C W Cl + ^ ^ 


6 -re • 6 0 

bh 


6 


6 


— T P bb r Q -f ~ 1 + Co dr o + C\dr l + C\dr Q I 

o L 5& J 


Let C = center height touched by diagonal 
H = side “ “ “ “ 

D = distance out to side height touched by diagonal 


S r — ^9 (^0 "1 br l *1 ^5) "1 C odr 0 + Ci^ + CyZ^rJ 

Then S,= |r|(/>,„ + h h + Hi) + Co di, + Cidi, + C,»<] 

= solidity for left half of section 
S=S r -\- S t = (hr 0 + hi Q + h Tl + hi t + H r + III) + Co(d»- 0 4- d; o ) 

+ Ci (dr x + dij + C,D r + CjZ^J (171) 


252. An example will further show the application of this 
method. 



The notes show the direction of the diagonal as taken on the 
ground. 


In this case 


b = 20 


s = 1 to 1 



















Methods of Computing Earthwork . 157 

253 . Henck gives note-book and calculations in this form : — 


Sta. 


h 

c 

hr 

d r 

di + d r 


D r C r 

DiCi 

1 

13.0 

3.0\ 

2.0\ 

1.0 

11.0 

24.0 

48.0 



0 

15.0 

54) 

\3.0 

\1.0 

11.0 

26.0 

78.0 

22.0 

39.0 


8.0 = 8.0 22.0 


4.0 80.0 

14.0 x — = 140.0 

2 v-- 

6 )827.0 


5450 (cu. ft.) 


254 . The calculations could be conveniently made however 
from the notes as now generally taken, as is shown below : 


1 

0 


13.0 

4- 

2.0 

11.0 

24.0 

X 

2.0 = 

48.0 

-f- 3.0s. 

15.0 \ 

-- “f- 

\ 

3.0 

+ 1.0 

11.0 

26.0 

X 

3.0 = 

78.0 

-1- 5.0 



+ 1.0 








2.0 

11.0 

X 

2.0 = 

22.0 

8.0 



8.0 

13.0 

X 

3.0 = 

39.0 


4.0 


14.0 x — = 140.0 

2 _ 

6)327.0 

5450 (cu. ft.) 


255 . The work of computation would not, in either of these 
cases, properly be done in the field note-book, but rather in 
a calculation book, or other suitable place. 

For a series of cross-sections, Henck systematizes the work, 
and reduces the labor noticeably from what is shown here. 
(See Henck’s Field Book.) Henck’s method is strictly accu¬ 
rate, upon the assumption made as to the upper surface. In 
general railroad practice, most engineers prefer to assume the 
upper surface a warped surface of the sort described. 

Henck’s method is less rapid than that of averaging end areas 
and applying the prismoidal correction. 
































158 Railroad Curves and Earthwork. 


256. The method of averaging end areas and applying the 
prismoidal correction appears in point of accuracy and rapidity 
to meet the requirements of modern railroad practice. 

Some engineers whose opinions are entitled to careful con¬ 
sideration object to the use of the prismoidal formula or 
prismoidal correction in any form, some as an unnecessary 
refinement, and some on the ground that certain practical con¬ 
siderations render the results nearer the truth when the method 
of averaging end areas is used without applying the prismoidal 
correction. Probably the greater part of the best engineering 
practice favors the use of the prismoidal correction. 


257-263. Example. 


Showing a comparison of various methods of calculating 
earthwork. 

Notes of excavation. Ease 24. Slope 1J to 1. 


Sta. 1 —^ + 

+ 1.0 


1.0 


16.5 
+ 3.0 


Sta. 0 


JL5_ + 10 .oJM- 

+ 11.0 + 21.0 


The mid-section will be 


Sta. 0+50-^ + 10.0 - r> ° ° 

+ 6.0 +12.0 


COMPUTING THE SOLIDITIES. 

I. Prismoidal Formula. 8 P = 39450 cu. ft. 

II. End Areas. S e — 45750 

Error + 6300 = 16 per cent. 

III. Middle Areas. S M = 36300 

Error — 3150 = 8 per cent. 

IV. Equivalent Level Sections. Si = 39370 

Error — 80 = 0.2 per cent. 

V. Mean Proportionals. S _= 36660 

Error — 2790 = 7 per cent. 

VI. Henck’s Method. 

One system of diagonals S h = 38100 

Diff. — 1350 = 3 per cent. 

Opposite system of diagonals S h = 40800 

Diff. + 1350 = 3 per cent. 

Mean shows no difference. 








CHAPTER XIII. 


SPECIAL PROBLEMS. 

264. Correction for Curvature. 

In the case of a curve, the ends of a section of earthwork are 
not parallel, hut are in each case normal to the curve. In cal¬ 
culating the solidity of a section of earthwork, we have hereto¬ 
fore assumed the ends parallel, and for curves this is equivalent 
to taking them perpendicular to the chord of the curve between 
the two stations. 

Then, as shown in Fig. 1 (where IG and GT are center-line 
chords), the solidity (as above) of the sections IG and GT will 
be too great by the wedge-shaped mass RGP, and too small by 


Q E S 



QGS. When the cross-sections on each side of the center 
are equal, these masses balance each other. When the cross- 
section on one side differs much in area from that on the other, 
the correction necessary may be considerable. 

159 







160 


Railroad Curves and Earthwork. 



In Fig. 2, use c, hi, h r , d t , d n b , s, as before. 

Let D — degree of curve. Make BL = AD, and join OL. 

Then ODAG balances 
OLBG, and there remains 
an unbalanced area OLE. 

Draw OKP parallel to 
AB. 

By the “ Theorem of 
Pappus’’(see Lanza, Ap¬ 
plied Mechanics), “If a plane 
area lying wholly on the same 
side of a straight line in its own 
plane revolves about that line, 
and thereby generates a solid 
of revolution, the volume of the 
solid thus generated is equal to 
the product of the revolving 
area and of the path described by the center of gravity of the 
plane area during the revolution.” 

The correction for curvature, or the solidity, developed by 
this triangle OLE (Fig. 2) revolving about OG as 
an axis will be its area x the distance described 
by its center of gravity. The distance 
out (horizontal) to the 
center of gravity from 
the axis (center line) 
will be two thirds of 
the mean of the dis¬ 
tances out to E and to 
L, or 

_ 2 di + d r 

~3 ’ 2 


Fig. 1. 



and the distance described will be 


| . i?L±ir x angle QGS 
The area OLE = OK x ^ 

2 















Special Problems. 

Therefore the correction for curvature, 


1G1 



dr + di 

3 


x angle QGS 


When 1G, GT are each a full station, or 100 ft. in length, 

QGS = D 


C = [5 -f • - r —- il • < k±Jk x angle D 
\2 / 2 3 

arc 1° = .01745 


0 = f- + sc > \ hr x (lr + — 1 x 0.01745 D 

VP I 2 3 

= Q + sc) (hr - hi)(dr + di) X 0.00291 D (cu. ft.) (172) 

= (| + sc) (/i r - A *)(dr + d{) x 0.00011 D (cu. yds.) (173) 


265. When IG or GT, or both, are less than 100 ft., let 
1G = lo and GT = li 

Then QGE= il x f andSGE = 4 x f 

Q GS =^ 

C = (l + sc) (h r - hi)(dr+di) x 0.00011 D (cu. yds.) (174) 


266. The correction C is to be added when the greater area 
is on the outside of the curve, and subtracted when the greater 

c area is on the inside of the curve. 
When the center height is 0, as 
in Fig. 3, we may consider this 
a regular section in which c = 0, 

7 

hi = 0, and di = -; then 
2 

C = |xA,x(rf, + |) x 0.00011 D (cu. yds.) (175) 













162 . Railroad Carves and RartfovorJc. 


In the case of an irregular section, as shown in Fig. 4, the 
area and distance to center of gravity (for example, of OHEML) 
may be found by any method available, and the correction 



Fig. 4. 

figured accordingly. The correction for curvature is, in present 
railroad practice, more frequently neglected than used. Never¬ 
theless, its amount is sufficient in many cases to fully warrant 
its use. 

267. Opening in Embankment. 

Where an opening is left in an embankment, there remains 
outside the regular sections the mass DEKHF. 




This must be calculated in 3 pieces, ADF, BEKH, ABHF. 


Let 


b — base = AB 

d r = distance out right 

d t = distance out left 

p r = BH l . 

_ | taken parallel to center line 

•£:} heights at 

Si = solidity ADF 
s 2 = BEKH 
s 3 = ABHF 



















Special Problems. 


163 


Then (approximately) following the “Theorem of Pappus,” 
Si = mean of triangular sections AD and AF x distance de¬ 
scribed by center of gravity. 


In the quarter cone AFD, AF = p t 

AD = — - 

2 

Then average radius R t = mh = ^ + ^ 

2 

Area of vertical triangular section A t = -—- 

2 

• 7? 

Distance from A to center of gravity of vertical section = — 

o 


ity _ B, v jr _ wBt 

x ^ (on. ft.) 


Arc described by center of gravity = — x — = 

6 J 3 2 0 


Si = 


fiR? x 3.1410 
2 x 6 x 27 


(cu. yds.) 


si = 0.0097 fiRi 2 (cu. yds.) 
Similarly, in the quarter cone BEKH 

The average radius R r = ^ - 


(170) 


fr Rr ^ Rr , 

s 2 = x -Q- ( cu - ft 0 


s 2 = 0.0097 f r R r 2 (cu. yds.) 


(177) 


For the solid AGBHF 


„ area AF 4- area BH , AD 

S 3 = - X Ad 

2 

+f-Pr)b 

4 


(178) 













164 


Railroad Curves and Earthwork. 


The work of deriving formulas (176) and (177) is approxi¬ 
mate throughout, but the total quantities involved are in gen¬ 
eral not large, and the error resulting would be unimportant. 

There seems to be no method of accurately computing this 
solidity which is adapted to general railroad practice. 

268. Borrow-Pits. 

In addition to the ordinary work of excavation and embank¬ 
ment for railroads, earth is often “borrowed” from outside 
the limits of the work proper; and in such excavations called 
“ borrow-pits,” it is common to prepare the work by dividing 
the surface into squares, rectangles, or triangles, taking levels 
at every corner upon the original surface ; again, after the 
excavation of the borrow-pit is completed, the points are repro¬ 
duced and levels taken a second time. The excavation is thus 
divided into a series of vertical prisms having square, rectangu¬ 
lar, or triangular cross-sections. These prisms are commonly 
truncated top and bottom. The lengths or altitudes of the 
vertical edges of these prisms are given by the difference in 
levels taken, 

1st, on the original surface, and 

2d, after the excavation is completed. 

This method of measurement is very generally used, and for 
many purposes. 

269. Truncated Triangular Prisms. 

Let A = area of right section EFD of a 
truncated prism, the base ABC 
being a right section 

hi = height AH 

h 2 =: “ BE 

h z = “ CK 

a = altitude of triangle EFD dropped 
from E to FD 

Let S = solidity of prism ABCKHE 
^ = “ “ “ ABCFDE 

s u = “ “ pyramid FDEHK 


K 







Special Problems. 


165 


Then 


si = A x AD = A x 


3 AD 

3 


A x ad + B E + CF 


s u = area DFKH x - 


O 

O 


x FD x - 

2 3 

- KF + HD . x FD x - 

2 


3 


_ KF -f HD 

3 


x A 


S = s l + s u = A ^^ + CF + ^^ 

= , 1 ( AD + HD)+ B E+(CF+KF) 

3 

_ ^ h\ + h% -f- h% 

3 


3 


(179) 


If the prism be truncated top and bottom, the same reason¬ 
in'^ holds and the same formula applies. 


270. Truncated Rectangular Prism. 


Let A — area of right section ABCD 
of a rectangular prism 
truncated on top (base 
is ABCD) 



hi = height AE 

h 2 - “ BG 

hs — “ KC 

h 4 = “ HD 

8 = solidity of prism 


b - AD = BC 


e — A8 = DC 


















166 


Railroad Curves and Earthwork. 


Then using method of end areas, 

AEHD_±j|GKC; x 0 
2 

, hi -f- h-i , 7 ^2 “E ^3 

- __ x a 

~ 2 

_ /ti -f h 2 + As + hi 
4 

£_ j. hi + h2 + h& +h± ( cu . f t .) (180) 

4 

4 . fti 4 -h 2 + h 3 + hi ( CU> yds .) (181) 

27 4 

We may find £, correct by the prismoidal formula, if we 
apply the prismoidal correction. The prismoidal correction 
C = 0, since D 0 - = 0 (or in this case AD - BC = 0). The 

formula therefore remains unchanged. It is evident from this, 
then, that the solution holds good, and the formula is correct, 
not only when the surface EH KG is a plane, but also when it is 
a warped surface generated by a right line moving always par¬ 
allel to the plane ADHE, and upon EG and HK as directrices. 

Some engineers prefer to cross-section in rectangles of base 
15' x 18'. In this case 

. * 

o _ 15' x 18' . h\ 4- h -2 + hs 4- lu ( cu yds.) 

.27 * 4 

= iQ fti + ftg + fta + ft j ( CU . yds.) (182) 

4 

Other convenient dimensions will suggest themselves, as 
10' X 13.5' or 20' x 13.5' or 20' X 27' 

By this method the computations are rendered slightly more 
convenient; but the size of the cross-section, and the shape, 
whether square or rectangular, should depend on the topog¬ 
raphy. The first essential is accuracy in results, the second 
is simplicity and economy in field-work, and ease of computation 
should be subordinate to both of these considerations. 












Special Problems. 


167 


271. Assembled Prisms. 

In the case of an assembly of prisms of equal base, it is not 
necessary to separately calculate each prism, but the solidity of 
a number of prisms may be calculated in one operation. 

In the prism B, 

g ^ U-2 -j~ U3 -f- 4 - &2 

4 

S = A ««+«4 , ±&4 + t 

c 4 

From inspection it will be seen, taking A as the common 
area of base of a single prism, and taking the sum of the 
solidities, that the heights a 2 , «s enter into the calculation of 


a 2 a 3 < 1 ,_a s 


b, 

B 

b 2 

C 

t> 3 

D 

b 4 

b 5 b 

E 

c, 

F 

c 2 

G 

C 3 

H 

c 4 

\ 

e 5 

- 

K 

d 3 

L 

M 

d 5 

N 



e 3 e* 


one prism only ; u 3 , « 4 into two prisms each; b u b s one only ; 
b 2 , b 5 into three prisms ; 6 8 , b 4 into four prisms; and similarly 
throughout. 


Let t\ — sum of heights common to one prism 


u 

44 

u 

« 

two prisms 

u 

44 

u 

u 

three “ 

u 

44 

a 

u 

four “ 


Then the total solidity, 

S t = A h+lh±lh±Ali (cu. ft.) (183) 

4 



t\ -f- 2 ty 4~ 8 ts -f- 4 ti 


(cu. yds.) 


4 


(184) 

















168 


Railroad Curves and Earthwork. 

272. Additional Heights. 

When the surface of the ground is rough it is not unusual 
to take additional heights, the use of which, in general, involves 
appreciable labor in computation, it being necessary commonly 
to divide the solid into triangular prisms, as suggested by the 
figures just below, which include the case of a trapezoid. 



The computations may be simplified in the two special cases 

which follow : 

(a ) When the additional height h c is 
in the center of the rectangle. 

Here the solid is composed of an 
assembly of 4 triangular prisms whose 

A 

right sections are of equal area = —• 

The solidity of the assembled prisms 

^ _ A 2 h\ -f- 2 h‘2 -f- 2 hs T 2 hi T 4 h c 

~J 3 

= — (2 h\ -f- 2 -f- 2 7*3 -f- 2 hi -f- 4 7i c ) 

12 

= —(3 hi T 3 7!&2 “I - 3 hs -f- 3 ^ 4 )+ - (4 h c — hi — h 2 — h$ — 7 * 4 ) 
12 12 

) ' . V ia 

g - | t>i 4- h<> -f h, 4- hi ^ A ^ hi -f ho 4- hs 4- hi ^j (185) 

or the total solidity is that due to the four corner heights plus 
the solidity of a pyramid of equal area of base and whose alti¬ 
tude is the difference between the center height and the mean 
of the four corner heights. 












Special Pi'oblems. 


169 


(5) When the additional height is at 
the middle of one side of the rectangle, h, 

6 = - • — (hi 4~ h m + hz + h% + hi + h m ) + 


o * 77 + ^4 + hz) 

o 2 



s — — (hi + h m 4 - hz + h<i 4 - hi 4* h m 4* 2 h m 4- 2 hi 4 - 2 h$) 
12 


— (hi 4- ^2 4- 3 ^3 4- 3A 4 4- 4 /t m ) 
12 


(3 hi 4 - 3 h<i 4- 3 hz 4" 3 hi) + (4 h m — 2 hi — 2 h 2 ) 

12 12 


hi 4 - h-i 4 - hz 4 - hi , A / ^ _ h\ 4 ~ ^ 2 ~\ 

4 3 V * 2/1 


(185 A) 


or the total solidity is that due to the four corner heights plus 
the solidity of a pyramid of equal area of base and whose alti¬ 
tude is the difference between the middle height and the mean 
of the adjacent side heights. 

Apparently the principle of the pyramid applies conveniently 
only in these two cases. 

For the case where the point lies on one of the sides, an 
alternate method of dividing the rectangle (or trapezoid) is 
indicated below. 



The details of the computation in this case need not be 
worked out here. 











CHAPTER XIV. 


EARTHWORK TABLES. 


273. The calculation of quantities can be much facilitated 
by the use of suitably arranged “ Earthwork Tables.” 

For regular “Three-Level Sections” very convenient tables 
can be calculated upon the following principles or formulas : — 


E 



Use notation as before for 

c, h t , h r , di, d r , s, l, A, S 
Then A = ABKL + OKE - ODL 

OK x EM OL x ND 


= c(b + sc) -f 
= c(b + sc) -f- 


^(EM - ND) 


= c(6 + sc) + 1Q + sc^ (h r — c - c A hi) 
A = c(b 4- sc) I 2 (2 SC ) (^ J P ~ 2 c) 


For a prism of base A and l = 50, 

aS" — 50 (cu. ft.) = — A (cu. yds.) 

27 


S = ^j C( ^ b + SC) + I(I +SC ) ( ^ f h r~ 2c ) (cu. yds.) (186) 


170 








Earthwork Tables. 


171 


274. For cross-sections of a given base and slope, that is, 
given b and s constant, we may calculate for successive values 
of c, and tabulate values of L and K as follows: — 



L 

K 

c 

— c(b + sc) 

27 v ' 

25 fb . \ 

— -+SC 

27 \2 J 


L represents the solidity for the level section. 

K is for use as a correction. The formula then adapts itself 
to this table for any desired values of c, hi, h r . 


S=L + K(]h + h r - 2c) (186 A) 

Having found for successive stations $ 0 and $i (each for a 
prism l = 50), then for the full section by “ end areas,” 


for 


$100 = $0 + $1 

Aq + A.\ 100 


$100 = 


2 


27 


50 Ap 50 Ai 
27 + 27 


$el00 = 


$0 + $1 


(187) 


275. When l is less than 100, 

S„ = (So + so (188) 

For level sections h t = h r = c 

hi -j- h r — 2 c — 0 

and the formula 

$ = L -f K(hi + h r — 2 c) 

becomes $ = L (189) 

for level sections, and the quantities for any given values of c 
can be directly taken from column L without any correction 
from column K. 

In preliminary estimates, or wherever center heights only 
are used, such tables are rapidly used. 













172 


Railroad Curves and Earthwork. 


276. Tables may be found at the back of the book, pages 
190, 191, calculated for 

1 . 6 = 20 s=l|tol 

2. b = 14 s = 1± to 1 


An example will illustrate their use, 

b = 14 s = li to 1 

Notes: — 


St a. 1 

13.0 
- 4.0 

-3.7 


12.4 

-3.6 

0 

10.6 

- 2.5 


10.3 

-2.4 


- 2.2 

Calculations : — 





3.7 L = 

134.0 

K= 11.6 

hi 

+ h r — 7.6 

+ 

2.3 

0 .2\ 


2 c = 7.4 


136.3 

2.32 


-- 0.2 

2.5 L = 

82.2 

K = 10.0 

hi 

■F hr — 4.6 

— 

4.0 

0.4^ 


2 c = 5.0 

$o = 

78.2 

4.00 


'^-0.4 


$100 = Si + So — 214.5 

277. There is also at the back of this book a “Table of 
Prismoidal Correction” calculated by the formula 

C' = 5 l i (co —cOCDo-Z)') 

3.7 =- 1.2 
25.4 = — 4.5 
1.39 

2.78( 10 

0.28 - 0.28 
C= 1.67 


In the example above 

c 0 — c.i = 2.5 — 

Do — D{ = 20.9 — 

From Table find opp. 4.5 for 1 

2 

2 _ 

10 

$ioo = $« — 214.5 
C= 1.7 
S p = 212.8 














173 


Earthwork Tables. 


278. When the section is less than 100 ft. in length, the 

l . 
100 ’ 

(190) 


prismoidal correction is made before multiplying by 

l 


that is. 


= {So + Si - C) 


100 


279. Tables based upon these formulas have been published 
as follows: — 

“ The Civil Engineer’s Excavation and Embankment Tables,” 
by Clarence Pullen and Charles C. Chandler, published by the 
“ J. M. W. Jones Stationery and Printing Co.,” Chicago. 

Tables are calculated for b = 12, 14, 16, 18, 20 

o _ i i in 

Tables of a similar kind, but calculated so that 

o _$i + .Vo 

£>ioo —--- 

are Hudson’s Tables, published by John Wiley & Sons, New 
York. 


280. For general calculation adapted both to regular “ Three- 
Level Sections” and to “Irregular Sections,” tables can be cal¬ 
culated upon the following principles and formulas : — 

These tables are, in effect, tables of “Triangular Prisms,” in 
which, having given (in feet) the base B and altitude a of any 
triangle, the tables give the solidity (in cu. yds.) for a prism of 
length l = 50 ; that is, 


^.W = W a B 

2 27 


(191) 


Whenever the calculation can be brought into the form 

S = — aB, the result can be taken directly from the table. 
54 


281. Tables of this kind are provided in Allen’s Field and 
Office Tables (XXXII) for bases 14 to 30, slope 1£ to 1. A 
sample page is shown at the end of this book. Convenient 
tables of the same kind, but arranged differently for use, are 
“Tables for the Computation of Railway and other Earth¬ 
work,” by C. L. Crandall, C. E., Ithaca, New York. 




174 


Railroad Curves and Earthwork. 


50 

282. In both tables the formula S = — aB takes form thus, 

50 54 

S = p-j- x width x height, and the tables are arranged as below. 



Heights. 

Widths 

50 

width x height 

54 


The application to “ Three-Level Sections ” is as follows : — 
We have formula (162), p. 142, 


A= (e + ±)D-» 
\ 2 s )2 4s 

and for a prism 50 ft. in length (l = 50) 


s= m a = w( c + b\ D _n m ± mb 

27 541 2 s 54 2 s 


(192) 


or S is the sum of two quantities, each of which is in proper 
form for the use of the tables. 

For cross-sections of a given base and slope (6 and s con¬ 
stants), — is a constant, and also — • — • b is constant. 

2 s 54 2 s 

We may then calculate once for all —, and call this B (a 
constant). 

Also and call this a constant E. 

54 2 a 


Then 


S = —(c+ B)D - E 
54 v ' 


and 


In using the tables, c + B — height 

1) = width 

As in the previous tables, having found So and *Si, 

$100 = So + $! 

I 


S t = {So + $i) 


100 









Earthwork Tables. 


175 


283. Example. Allen’s Tables for Earthwork Computation. 


Notes: — 

Sta. 1 

9 -L |? 

7.3 

- 2.4 

- 1.2 

Sta. 0 

r— 

o 

1 

CO CM 

OO CM 

1 

6.4 

- 0.6 

6 = 11 

s = 1£ to 1 



cq 

II 

CO 

II 

-US 


Grade triangle, 

— x 3.7 x 11 

54 



Under height 3.7, find 1 = 3.43 

1 = 3.43 


Station 1. c = 1.2 
B= 3.7 

height = 4.0 

D- 9.1 + 7.3 = 16.4 

Under height 4.9, find 1 = 4.54 

6 = 27.22 
4 = 18.15 


10. =34.3 
1. = 3.4 

E = 37.7 


10. =45.4 
6 . =27.2 
.4= 1.8 

74.4 
E = 37.7 


Station 0. c — 0.7 
B = 3.7 

height = 4,4 


Si = 36.7 


]) = 8.8 -f 6.4 = 15.2 
Under height 4.4, find 


1 = 4.07 
5 = 20.37 
2= 8.15 


10. = 40.7 
5. =20.4 
. 2 = 0.8 

61.9 
£ = 37.7 


So = 24.2 
o = Si + So — 60.9 










176 Railroad Curves and Earthwork. 

284. Irregular Sections. 



An “ Irregular Section ” can be divided into triangular parts, 
as in the figure. Taking generally two triangular parts together 
for purposes of calculation, we have 


A x - h * x ( AG ~ ^h) 
2 

Ao _ x C (l i ~ f h) 

2 

A s - h ' x ~ 

2 

^ _ c x (d ( + d L ) 

2 

As x (d P - 0) 

2 

4 _ Ap x (d B — d L ) 

-^o — -■ 

2 

A B x (d r — d P ) 

2 


si=^VAG-r? H ) 

54 

82 = !% (<?<-<*,) 
54 

50 , , 

S3 = — Mh 
54 


«4 = —c(d, + d L ) 
0 50, 7 

§5 — —h L rtp 


Sr, — 77 Ap(dg — d L ) 
54 

s^ = ^A B (d r — dp) 


S — Si + S2 + S3 S4 + S5 -f Sq 4 ' S7 

S100 = So + Si 


•193) 


S, = (So + Si) 


l 

100 


285. The calculation of Irregular Sections in rough couhtry 
becomes very laborious unless the best methods are used, and 
this process should be thoroughly understood, 












CHAPTER XV. 


EARTHWORK DIAGRAMS. 


286. Computations of earthwork may also be made by means 
of diagrams from which results may be read by inspection 
merely. 

The principle of their construction is explained as follows : — 
Given an equation containing three variable quantities as 


x = zy 


(194) 


If we assume some value of z (making z a constant), the 
equation then becomes the equation of a right line. 

If this line be platted, using rectangular coordinates (as the 
line z — 1 in the figure), then having given any value of y, the 

corresponding value of x may be 
taken off by scale. If a new value 
of z be assumed, the equation is 
obtained of a new line which may 
also be platted (as z — \ in the 
figure), and from which also, hav¬ 
ing given any value of y , the cor¬ 
responding value of x may be 
determined by scale. Assuming 
a series of values of z and platting, 
we have a series of lines, each 
representing a different value of z, 
and from any one of which, having given a value of y, we may 
by scale determine the value of x. 

Thus, given, values of z and y ; required , x, we may find, 



1 . The line corresponding to the given value of 2 !, and 

2. Upon this line we may find the value of x corresponding 
to the given value of y. 


177 





178 Railroad Curves and Earthwork. 


287. Next, if instead of platting upon lines as coordinate 

axes, we plat upon cross-section 
paper, the cross-section lines form 
a scale, so that the values of x and 
y need not be scaled, but may be 
read by simple inspection as in the 
figure. 


288. If the equation be in'the 
form 

x = azy (195) 

the same procedure is equally possible, and the line represent¬ 
ing any value of z will still be a right line. 

If the equation be in the form 

x — a{z -f- b) (y + c) -f d (196) 

in which a, b, c, d, are constants, the same procedure is still 
possible, and the line representing a given value of z is a right 
line, as before. 

The use of diagrams of this sort is therefore possible for the 
solution of equations in the form of 

x = a(z + b)(y + c) + d 

or in simpler modifications of this form. 

289. Referring again to the figure above, we may consider 
the horizontal lines to represent successive values of x and refer 
to them as the lines 

£ = 0; x — 1; x — 2, etc. 

and similarly we may refer to vertical lines as the lines 

y = 0; y = 1 •, y = 2, etc. 
just as we refer to the inclined lines 

2 = z ~ 1, etc. 

Having given any two of the quantities x, y, z , the third may 
be found by inspection from the diagram by a process similar 
to that described. 





















Earthwork Diagrams. 179 

290 Diagram for Prismoidal Correction. 

Formula O = -1- (c 0 - c x ) ( D 0 - Di) (168) 

o.24 

This lias the form x = a x £ x 

Construction of diagram. 

Assume (as we did for z) a series of values of 

c 0 — Ci = 0, 1, 2, 3, 4, 5, etc. 

When co — Ci = 0 then 0=0 
or, the line Co — Ci coincides with the line 0 = 0. 

When Co — ci = 1, the equation of the line Co — Ci is 

O = — (Do - DO 
3.24 v 

To plat this right line, we must find two or more points on 
the line. For the reason that cross-section paper is generally 
warped somewhat, it is best to take a number of points not 
more than 3 or 4 inches apart, in order to get the lines suffi¬ 
ciently exact. For convenience, take values of Do — D\ as 
follows : — 

When (co — Ci) = 1 

take Do — D\ — 0, 3.24, 6.48, 9.72, 12.96, 16.20, etc. 

then C — 0 , 1 , 2 , 3, 4, 5, etc. 

When co — ci = 2, the equation of the line c 0 — Ci is 


Therefore when Co — Ci = 2 


take 

o 

• 11 

1 

o 

3.24, 

6.48, 

9.72, 

12.96, 

16.20, etc. 

then 

0 = 0 , 

2 , 

4 , 

6 , 

8 , 

10 , etc. 





180 Railroad Carves and Earthwork. 


291. In like manner a table may be constructed. 



0 

3.24 

0.4S 

9.72 

12.90 

10.20 

19.44 

22. OS 

20.92 

A) - A 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 


1 

0 

1 

2 

3 

4 

5 

6 

7 

8 


2 

0 

2 

4 

6 

8 

10 

12 

14 

16 


3 

0 

3 

6 

9 

12 

15 

18 

21 

24 


4 

0 

4 

8 

12 

16 

20 

24 

28 

32 


5 

0 

5 

10 

15 

20 

25 

30 

35 

40 


6 

0 

6 

12 

18 

24 

30 

36 

42 

48 


7 

0 

7 

14 

21 

28 

35 

42 

49 

56 


8 

0 

8 

16 

24 

32 

40 

48 

56 

64 


9 

0 

9 

18 

27 

36 

45 

54 

63 

72 


10 

0 

10 

20 

30 

40 

50 

60 

70 

80 


c- — c. 












292. It will be noticed that when Do — Di = 0 , C = 0. 

Therefore for all values of c 0 — Ci, the lines pass through the 
origin. 

We may proceed to plat the lines co — Ci — 1, Co — C\ = 2, 
c 0 — Ci = 3, etc., from data shown in the above table, platting 
upon the lines Do — D i = 3.24, Do — D\ — 6.48, etc., the points 
shown with circles around them in the cross-section sheet, 
p. 181. 

Having the lines c 0 — c x = 1, c 0 - c x = 2, 3, platted, inter¬ 
mediate lines are interpolated mechanically upon the prin¬ 
ciple that vertical lines would be proportionally divided (as 
ML is proportionally divided into 5 equal parts), and points 
are marked for the lines 

c 0 -d = 1.2, 1.4, 1.6, 1.8 

For the most convenient use, the values of c 0 — Ci are taken 
to every second tenth of a foot in interpolating, as is shown on 
the diagram, p. 181, between 1 and 2 ; that is, 

1.2, 1.4, 1 . 6 , 1.8 

A complete diagram is shown at the back of the book. 







































jE WrthworJc Diagrams 


181 














































































































































182 Railroad Curves and Earthwork. 


293. For Use. 

Find the diagonal line corresponding to the given value of 
c 0 — Ci ; follow this up until the vertical line representing the 
given value of Z>o — Di is reached, and the intersection is thus 
found. Then read off the value of <7 corresponding to this 
intersection. 


Example. c 0 — Ci = 1.2 

Do — Di = 11.0 

again, c<> — ci =1.7 

Do - D x - 7.0 


G = 4.0 
C = 3.6 ± 


294. Diagram for Triangular Prisms. 

50 

From formula (191), S = — cD, a table may be constructed. 



0 

5.4 

10.8 

16.2 

21.6 

27.0 

I ) 

0 

0 

0 

0 

0 

0 

0 


1 

0 

5 

10 

15 

20 

25 


2 

0 

10 

20 

30 

40 

50 


3 

0 

15 

30 

45 

60 

75 


4 

0 

20 

40 

60 

80 

100 

I 

5 

0 

25 

50 

75 

100 

125 


6 

0 

30 

60 

90 

120 

150 


7 

0 

35 

70 

105 

140 

175 


8 

0 

40 

80 

120 

160 

200 


9 

0 

45 

90 

135 

180 

225 


10 

0 

50 

100 

150 

200 

250 


c 


• 







From this a diagram can be constructed similar in form to 
that for Prismoidal Correction. 

The lines for all values of c pass through the origin. 

In constructing this table, any values of D might have been 
taken instead of those used here. Those used were selected 
because they give results simple in value, easily obtained, and 
readily platted. 

























Earthwork Diagrams. 183 

295. Diagram for Three-Level Sections. 

Formula, 8 = ^ (c 4- D — ~ • b ( 192 ) 

54 \ 2 s) 54 2 s 

A separate diagram will be required for each value (or com¬ 
bination of values) of b and s. Since b and s thus become 
constants, the formula assumes the form of 

x = a(z + b)y + d (197) 

and the diagram will consist of a series of right lines. 

A table can be made up by taking successive values of c = 0, 
1, 2, 3, 4, etc., and finding for each of these the value of 8 
corresponding to different values of Z), using the above formula. 

To make separate and complete computations directly by 
formula would be quite laborious ; there is, however, a method 
of systematizing the construction of the table which can be 
shown better by example than in any other way. 


296. Example. 
Formula 

becomes 


14 


s = 1 \ to 1 


r.- h 


S = ^(c + —\D-Q0Ad 
54 \ 3 ) 


(198) 


A table has been prepared for successive values of 

c = 0, 1, 2, 3, 4, 5, etc. 

and for D = 14, 10.2, 21.6, 27.0, etc. 

These values of D are selected for the following reasons: 
I) = 14 is the least possible value ; D = 16.2, 21.6 are desirable 
because they are multiples of 5.4, and th e# factors in the for¬ 
mula show that the computations will be simplified by selecting 
multiples of 5.4 for the successive values of D. 


184 


Railroad Curves and Earthwork 



14 

16.2 

21.6 

27.0 

32.4 

37.8 

43.2 

D 


12.963 

15. 

20. 

25. 

30. 

35. 

40. 

Const. 

diff. 

0 

0 

9.51 

32.84 

56.18 

79.51 

102.84 

126.18 


1 

12.963 

24.51 

52.84 

81.18 

109.51 

137.84 

166.18 


2 

25.926 

39.51 

72.84 

106.18 

139.51 

172.84 

206.18 


3 

38.889 

54.51 

92.84 

131.18 

169.51 

207.84 

246.18 


4 

51.852 

69.51 

112.84 

156.18 

199.51 

242.84 

286.18 


5 

64.815 

84.51 

132.84 

181.18 

229.51 

277.84 

326.18 


6 

77.778 

99.51 

152.84 

206.18 

259.51 

312.84 

366.18 


7 

90.741 

114.51 

172.84 

231.18 

289.51 

347.84 

406.18 


8 

103.704 

129.51 

192.84 

256.18 

319.51 

382.84 

446.18 


9 

116.667 

144.51 

212.84 

281.18 

349.51 

417.84 

486.18 


10 

129.630 

159.51 

232.84 

306.18 

379.51 

452.84 

526.18 


c 










When 

c = 0 


- 60.49 

When 

D = 14 

s =®-¥-14 

- 60.49 

When 

D = 16.2 

= 00.49 

- 60.49 


we may again calculate directly 

8 = f£ . if . 16.2 - 60.49 

but a better method is to find how much greater 8 will be for 
D = 10.2 than for D — 14.0. 

We have 8 = f£ • • D - 60.49 * 

Then for any new value D' 

S' = |£ . • 2>' - 60.49 

( 199 ) 

for /)' = 16.2 D = 14.0 D 1 - D = 2.2 

- >9 = |f • V- x 2.2 = 9.51 
*S = 0 

=9.51, which is entered in table. 






































Earthwork Diagrams . 


185 


Similarly, S" - S' = f • VKZ>" - Z>') 

= 21.6 Z)' = 16.2 D" - D' = 5.4 


Similarly, 


/S'" - £' = ff x - 1 / x 5.4 


= 23.3:53 
S' — 9.51 

S" = 32.843 
S’" - S" = 23.333 

S'" = 56.176 
S iv - S"' = 23.333 


S w = 79.509 


S w = 79.509 
23.333 

S v = 102.842 
23.333 

S vi = 126.175 


Constant increment for D'— D = 5.4 is 23.333. 


297. Each result is entered in the table in its proper place. 
The final result for c = 0 and D = 43.2 should be calculated 

independently as a check. 

When c = 0 S = $£ • ^ • D — 60.49 

When D = 43.2 S = f£ • x 43.2 - 60.49 

= 50 x - 1 / x 0.8 - 60.49 

= - 60.49 

= 186.67 - 60.49 

S= 126.18 

This checks exactly, and all intermediate values are checked 
by this process, which is also more rapid than an independent 
calculation for each value of D. 

298. We now have values of S for the various values of 
D = 14.0, 16.2, 21.6, etc., when c = 0. 

Next, find how much these will be increased when c — 1. 

Formula S — ff (c -f — 60.49 

for any new value c f S' — (c’ + y)Z> — 60.49 

S'-S=M(c' -c)D (200) 







186 


Railroad Carves and Earthivork. 


When c' = 1 and c = 0, c r — c = 1 

S , -8 = W D 

Similarly, S" - 8' = “ C ’) D 

When c" = 2 and c' = 1, c" — c' = 1 

S"-S’ = $$D 

That is, for any increase of 1 ft. in the value of c, 

S’-S = $iD (201) 

When D= 14 

y-«=f| X 14 = 12.963 

This we enter as the constant difference for column D = 14. 

We have already found So = 0 

12 963 
S\ = 12.963 
12.963 
S 2 = 25.926 

This gives column 14. Ss = 38.889 etc. 

When D = 16.2 

( 201 ) & - S = X 16.2 = 50 x 0.3 

= 15 

Enter 15 as the constant difference in column 16.2. 

We already have S 0 = 9.51 

15. 

Si = 24.51 
15. 

S> - 39.51 

This allows us to complete column 16.2. £3 = 54.51 etc. 

Similarly for D — 21.6 S' — S — 20 

Enter 20 as constant difference in column 21.6, and complete 
column as shown in table. 

Similarly, fill out all the columns shown in the table. 






Earthwork Diagrams. 


187 


299. The final result for c = 10, D = 43.2 should be calcu¬ 
lated independently, and directly from the formula, as a check. 


S=™ (c + *£)D -00.49 

c = 10 D = 43.2 

= 14.607 x 43.2 - 60.49 

= 50 x 14.607 x, 0.8 - 60.49 
= 40 x 14.66 7 - 60.49 

= 586.68 - 60.49 

S = 526.19 


The table gives 526.18. This checks sufficiently close to indi¬ 
cate that no error has been made. It would yield an exact 
check if we took c + J £ — 14.6667. 


300. Note that for 

o 

H 

II 

D = 43.2 

value 

= 526.18 


c = 10 

D = 37.8 

66 

452.84 




Diff. 

= 73.34 

Between 

and 

o o 

I—1 T—1 

II II 

D = 37.8 \ 
D = 32.4 i 

Diff. 

= 73.33 


In line c = 10 a constant difference is found between succes¬ 
sive values of D differing by 5.4. This may be demonstrated 
to be = 73.33. 

All values in the table except column 14 are satisfactorily 
checked by applying this difference of 73.33 in line 10 together 
with the independent check of c = 10, D = 43.2. 

The value of c = 10, D = 14 can also be checked and shown 

to be correct. 

301. Having the table, page 184, completed, the construction 
of the diagram is simple. 

The “Diagram fur Three-Level Sections , Base 14, Slope \\ 
to 1„” was calculated and constructed according to this table. 
The Diagram given shows a general arrangement of lines and 
figures convenient for use. For rapid and convenient use, the 
diagram should be constructed upon cross-section paper, Plate 
G ; and in this case the diagram will be upon a scale twice that 
of the diagram accompanying these notes. 



188 


Railroad Curves and Rarthivork . 


A “ curve of level section ” has been platted on this diagram 
in the following manner. For level sections, when 

c = 0 D = 14.0 c = 2 D = 20.0 

c = 1 D= 17.0 c = 0 D = 32.0 

c = 1.4 I) = 18.2 etc. 

The line passing through these points gives the “curve of 
level section.” 

Aside from the direct use of this curve of level section (for 
preliminary estimates or otherwise), it is very useful in tending 
to prevent any gross errors in the use of the table, since, in 
general, the points (intersections) used in the diagram will lie 
not far from the curve of level section. 

302. Use of Diagram. 

Find the diagonal line corresponding to the given value of c; 
follow this up until the vertical line representing the given value 
of D is reached, and this intersection found. Then read off the 
value of 8 corresponding to this intersection. 


Example . Notes. 



Sta. 1 

13.0 

-4.0 

-3.7 ,2 ' 4 

— 343 

St = 136. 

Sta. 0 

10.6 

o k 10.3 

II 

—j 

00 

• 

-2.4 

-2.2 


S =214. 

For Sta. 1 

c = 3.7 

D = 25.4 



c = 3.7 is the middle of the space between 3.6 and 3.8. 

Follow this up until the vertical line 25.4 is reached. 

The intersection lies upon the line S\ = 13G. 

Enter this above opposite Sta. 1. 

For Sta. 0 c = 2.5 D = 20.9 

c = 2.5 is the middle of space between 2.4 and 2.6. 

Follow this up until the middle of space between 20.8 and 
21.0 is reached. 

The intersection lies just above the line 
8 0 = 78 

Enter this opposite Sta. 0. 

£ioo — St + So 

— 136 + 78 = 214 cu. yds. 







Earthwork Diagrams. 


189 


The prismoidal correction may be applied if desired. 

It should be noticed that in each case the intersection was 
quite close to the “ curve of level section.” 

303. Diagrams may be constructed in this way that will 
give results to a greater degree of precision than is warranted 
by the precision reached in taking the measurements on the 
ground. 

In point of rapidity diagrams are much more rapid than tables 
for the computation of Three-Level Sections. 

For “ Triangular Prisms ” and for Prismoidal Correction , 
the diagrams are somewhat more rapid. 

For Level Sections , the tables for Three Level-Sections, § 274, 
are at least equally rapid. 

A book entitled “Computation from Diagrams of Railway 
Earthwork,” by Arthur M. Wellington, published by D. Apple- 
ton & Co., N.Y., explains the application and construction of 
certain other tables in addition to those given here. “ Welling¬ 
ton’s Diagrams,” as there published, are upon a scale differing 
from that used here, and they do not allow of as great precision, 
but, on the other hand, are arranged to cover a large number of 
tables differing somewhat as to base and slope. 

304. The use of approximate methods for applying the pris¬ 
moidal correction to irregular sections (§§ 244-245) will be 
rendered practicable by the use of these “ Diagrams for Ihree- 
Level Sections.” 

Method 1. No use of diagrams is necessary. 

Method 2. Having found for any irregular sections (by tri¬ 
angular prisms or any other method) the solidity S for 50 ft. 
in length, find upon the diagram the line corresponding to 
this value of S; follow this line to the curve of level section, 
and read off the value of c (for a level section) which corre¬ 
sponds, and also the value of D for the same section. 

Method 3. Having found in any way the value of S ; if c is 
given, find the value of D to correspond ; if D is given, find the 

value of c to correspond. 

Method 4. The use of diagrams is not needed. 


CHAPTER XVI. 


HAUL. 

305. When material from excavation is hauled to be placed 
in embankment, it is customary to pay to the contractor a 
certain sum for every cubic yard hauled. Oftentimes it is pro¬ 
vided that no payment shall be made for material hauled less 
than a specified distance. In the east a common limit of “free 
haul” is 1000 ft. Often in the west 500 ft. is the limit of 
“free haul.” Sometimes 100 ft. is the limit: 

A common custom is to make the unit for payment of haul, 
one yard hauled 100 ft. ; the price paid will often be from 1 to 
2 cents per cubic yard hauled 100 ft. 

The price paid for “ haul ” is small, and therefore the stand¬ 
ard of precision in calculation need not be quite as fine as in 
the calculation of the quantities of earthwork. The total 
“haul” will be the product of 

(1) the total amount of excavation hauled, and 

(2) the average length of haul. 

306. The average length of haul is the distance between the 
center of gravity of the material as found in excavation, and 
the center of gravity as deposited. It would not, in general, be 
simple to find the center of gravity of the entire mass of exca¬ 
vation hauled, and the most convenient way is to take each 
section of earthwork by itself. The “haul” for each section 
is the product of the 

(1) number of cubic yards in that section, and 

(2) distance between the center of gravity in excavation, 

and the center of gravity as deposited. 

100 


Haul. 


191 


307. When excavation is placed in embankment, there may 
be some difficulty in determining just where any given section 
of excavation will be placed, and where its center of gravity 
will be in embankment. 



In hauling excavation in embankment, there is some plane, 
as indicated by AB, to which all excavation must be hauled on 
its way to be placed in embankment, and (another way of put¬ 
ting it) from which all material placed in embankment must 
be hauled on its way from excavation. We may figure the 
total haul as the sum of 

(1) total “haul” of excavation to AB, and 

(2) total “haul” of embankment from AB. 

The total “haul” of excavation to AB and the total “haul ” 
of embankment from AB will most conveniently be calculated 
as the sum of the hauls of the several sections of earthwork. 
For each section the haul is the product of 

(1) the solidity S of that section, and 

(2) distance from center of gravity of that section to the 

plane AB. 

308. When the two end areas are equal, the center of gravity 
will be midway between the two end planes. When the two end 
areas are not equal in value, the center of gravity of the section 
will be at a certain distance from the mid-section, as shown 
by the formula 

„ _ l 2 - A 0 

g 12 S 

in which x g = distance center of gravity from mid-section. 














192 


Railroad Curves and Earthwork . 


309. Referring to the figure below, and following the same 
general method of demonstration used previously, § 235, 


D, 



let bo = base = A 0 B 0 

bi- “ =AxBi 

Co = center ht. = 0 0 Go 
Ci = center ht. = OiGi 
l = length (altitude) 
of section=G 0 Gi 

Aq — area of D 0 A 0 BoEo 
A\ = area of D 1 A 1 B 1 E 1 
8 = solidity 


Also use notation b x , c x , A x for a section distant x from Gi. 
Find the distance of the center of gravity from A 1 B 1 E 1 D 1 , and 
let this = x c . Let x g = distance of center of gravity from mid¬ 
section. 

Then for any elementary section of thickness dx and distance 
X from AiBiEiDi, its moment will be 


[y>i + (b 0 - bx) ^ci + (c 0 - ci) x dx 
S'X e =abi +(&o - &i)^j [ci + (Co - Cl) *J x dx 


S-Xc 


bjCjP , 6i(co-cP/» , Cj(bo-bi)l s , (cq-Ci)(5 0 -5i)Z 4 


2 


3 l 


3 l 


4Z2 


11 

12 


6 61C1 + 4 Z>iCo + 4 5 0 ci + 3 boCo 

— 4 b\Ci — 3 6ic 0 — 3 &0C1 

- 4 &1C1 
+ 3 b\C\ 


S’ x c = 


x c = 


V- 

i<> * (^i c i t>\Co + &0C1 + 3 5 0 Co) 

— x d~ bjCo 4 - 5 pCi + 3 bpCp 

12 S 


( 202 ) 













Haul 


193 


Wliat is wanted is x g rather than x c . 


* g = g — Xc 


Sx g — S - — Sx c 
2 

S = - (2 61 C 1 -f- 2 boCo + biCo + 6 qCi) 

o 

$ • — —(2 & 1 .C 1 + 2 6 0 Co + &i c o + &o<h) 

J Id 

72 

/S' • £ c = ~ (5iCi + 3 6(A) ~f &i c o + 6oCi) 

1Z 

72 

/S' • % = — (&iCi - t 0 C 0 ) 

liJ 


= Ii 




x a = ~ A 1 — (/S' in cu. ft.) 


— 


12 /S' 

? 2 - ^4q 

12 x 27 * /S’ 


(/S in cu. yds.) 


(161) 


(203) 


/ 


(204) 

(205) 


310. The formula above applies to the solid shown in the 
figure, wdiich has trapezoidal ends, but it will apply also when 
D 0 Ao, DiAi, are each =0, and therefore applies to-such solids 
having triangular ends ; and since any section of earthwork 
with parallel ends may be divided into a number of such solids 
with triangular ends, it applies to all ordinary sections of rail¬ 
road earthwork, since it applies to the parts of which it is 
made up. 

To show that in fact this formula is correct for prisms, wedges, 
and pyramids, use a method similar to that shown on page 145 ; 
find for each solid an expression for x g in terms of A and i ; 
reduce to the form 

_ A\ — A,) 

S 






194 Railroad Curves and Earthwork . 

311. The formula 

= 7:2 

9 12 x 27 $ 


is not in form convenient for use, because we have not found 
the values of Ai and Ao, but instead have calculated directly 
from the tables or diagrams, the values of Si and So for 50 ft. 
in length, where 


$i = ~ -A u or A\ 


27 Si 
50 


and 




Substituting, 


. _ 100 x 100 $i - So 27 
9100 12 x 27 S ' 50 


_100 Si - So 

3100 6 S 


(206) 


This formula is in shape convenient for use, and results 
correct to the nearest foot can be calculated with rapidity. 


312. For a section of length l less than 100 ft. 


1 2 

Xn. = - 

91 12 x 27 


Ai — Ap 
Si 


l 2 


_ A\ — Ap 

12x27 $1.00 X —— 

100 


100 l 
12 x 27 


Ai — Aq 


$ 


100 


Xn... = 


100 x 100 A j - Ao 


9m 12 x 27 


$100 


X 9i = X 


■9100 * 


100 


( 207 ) 



















Hauh 


105 


313. It is not, however, always necessary to calculate the 
position of the center of gravity of each station, or to calculate 
for each station the correction x g . It may often be easy to 
calculate for a series of sections a correction to be applied 
to obtain the correction in haul for the entire mass. 



To find the correction in haul for the entire mass, let 


X c = cent, of grav. for entire mass (approximately), 



„ A , . r Total Haul 

A = true dist. to r.g. of entire mass =--- 

8 

Xg = X- X C 

S 0 = A 0 , Si = Ai, S -2 = etc., as taken from tables 

or diagrams. 


When all sections are of uniform length, 100' as in figure 
above, 


Approximate total haul 
X c S = f(S a 
Correct total haul 

XS=S a (l + X g a 


+ 3 S b 


+ 5 SJ 


S(X-X C )= SaXga 

100 




6 L 


Sa 


+ Sb 4- Xgh^J 4* Sc 0 - + Xgc^j 

+ ShXgb + ScXgc 

. O So — /S's'l 
+ 


Si - S-2 


SXa = 


100 


(So - Sz) 


Or in general 

SX g = ( S 0 — S n ) = correction in total haul. (208) 













CHAPTER XVII. 


MASS DIAGRAM. 

314. Many questions of “ haul ” may be very usefully treated 
by means of a graphical method, known to some as “Mass 
Leveling,” in which is used a diagram sometimes called a 
“Mass Profile,” but which will be referred to here as the 
“Mass Diagram.” 

The construction of the “ Mass Diagram ” will he more clearly 
understood from an example than from a general description. 

Let us consider the earthwork shown by the profile on p. 198, 
consisting of alternate “cut” and “fill.” To show the work 
of constructing the “diagram ” in full, the quantities are calcu¬ 
lated throughout, but for convenience and simplicity, “level 
sections” are used and prismoidal correction disregarded. In 
a case in actual practice, the solidities will have been calculated 
for the actual notes taken. 

315. In the table, p. 197, the 

1 st column gives the station. 

2 d column gives center heights. 

3d column gives values of S from tables. 

4th column gives values of $100 or Si for each section, and 
with sign + for cut or - for fill. 

5th column gives the total, or the sum of solidities up to each 
station ; and in getting this total, each -f solidity is added and 

each — solidity is subtracted, as appears in the table from the 
results obtained. 

Having completed the table, the next step is the construction 
of the “ Mass Diagram,” page 198. In the figure shown there, 
each station line is projected down, and the value from column 
5, corresponding to each station, is platted to scale as an offset 
from the base line at that station, all -f quantities above the 
line, and all - quantities below the line. The points thus found 
are joined, and the result is the “ Mass Diagram.” 

190 


Mass Diagram 


197 


Station. 

Center 

Heights. 

Solidity for 
50' DUE TO 
Center 
Height given 
(Taken from 
Tables). 

Solidity 

for 

Section. 

Solidity 

Totals. 

0 

0 

0 



0 

1 

+ 1.7 

71 

+ 71 

+ 

71 

2 

+ 2.7 

120 

+ 191 

+ 

262 

3 

0 

0 

+ 120 

+ 

382 

4 

-3.3 

116 

- 116 

+ 

266 

5 

-5.1 

204 

-320 

— 

54 

6 

-2.9 

99 

I 

CO 

o 

CO 

— 

357 

7 

0 

0 

- 99 

— 

456 

8 

+ 2.4 

105 

+ 105 

— 

351 

9 

+ 4.5 

223 

+ 328 

— 

23 

10 

+ 2.5 

110 

+ 333 

+ 

310 

11 

0 

0 

+ 110 

+ 

420 

12 

-3.0 

103 

- 103 

+ 

317 

13 

-5.3 

215 

- 318 

— 

1 

14 

- 7.6 

357 

- 572 

— 

573 

15 

-8.4 

414 

-771 

— 

1344 

16 

-4.3 

163 

-577 

— 

1921 

17 

0 

0 

- 163 

— 

2084 

18 

+ 2.6 

115 

+ 115 

— 

1969 

19 

+ 3.6 

169 

+ 284 

— 

1685 

20 

+ 4.9 

248 

+ 417 

— 

1268 

21 

+ 6.7 

373 

+ 621 

— 

647 

22 

+ 7.5 

434 

+ 807 

+ 

160 

23 

+ 5.2 

268 

+ 702 

+ 

862 

24 

+ 2.4 

105 

+ 373 

+ 

1235 

25 

0 

0 

+ 105 

+ 

1340 

26 

-3.6 

129 

- 129 

+ 

1211 

27 

-6.0 

256 

-385 

+ 

826 

28 

-5.0 

199 

-455 

+ 

371 

29 

-2.6 

86 

-285 

+ 

86 

30 

0 

0 

- 86 


0 














198 


Railroad Curves and Ea) tliwork. 

















































































































































Mass Diagram. 


199 


316. It will follow, from the methods of calculation and con¬ 
struction used, that the “ Mass Diagram” will have the follow¬ 
ing properties, which can be understood by reference to the 
profile and diagram, page 198. 

1. Grade points of the profile correspond to maximum and 
minimum points of the diagram. 

2. In the diagram, ascending lines mark excavation, and de¬ 
scending lines embankment. 

:l. The difference in length between any two vertical ordinates 
of the diagram is the solidity between the points (stations) at 

which the ordinates are erected. 

4. Between any two points where the diagram is intersected 
by any horizontal line, excavation- equals embankment. 

5. The area cut off by any horizontal line is the measure of 
the “ haul ” between the two points cut by that line. 


317. It may be necessary to explain the latter point at some¬ 
what greater length. 

Any quantity (such as dimension, weight, or volume) is often 
represented graphically by a line ; in a similar way, the product 
of two quantities (such as volume into distance, or as foot 
pounds) may be represented or measured by an area. In the 
case of a figure other than a rectangle, the value, or product 
measured by this area, may be found by cutting up the area by 
lines, and these lines may be vertical lines representing volumes 
or horizontal lines representing distance. The result will be 
the same in either case. An example will illustrate. 

In the two figures let 

a and b represent pounds 
c “ feet 

b and the area of the trapezoid 
represent a certain number 
of foot pounds. The trape¬ 
zoid may be resolved into rectangles by the use of a vertical 
line, as shown in Fig. 1, or by a horizontal line, as in Fig. 2. 

In Fig. 1, the area is ax--i-b x - 



In Fig. 2, the area is 


a + b 
2 


X c 


the result of course being the snmo in both cases. 








200 


Railroad Curves and Earthwork. 








































































Mass Diagram. 


201 


318. In an entirely similar way, the area ABC (p. 200) repre¬ 
sents the “haul” of earthwork (in cu. yds. moved 100 ft.) 
between A and C, and this area may be calculated by dividing 
it by a series of vertical lines representing solidities, as shown 
above G and F. That this area represents the haul between 
A and C may be shown as follows : — 

Take any elementary solidity dS at D. Project this down 
upon the diagram at F, and draw the horizontal lines FG. 

Between the points F and G (or between D and I), there¬ 
fore, excavation equals embankment, and the mass dS must be 
hauled a distance FG, and the amount of “haul” on dS will 
be dS x FG, measured by the trapezoid FG. Similarly with any 
other elementary dS. 

The total “haul” between A and C will be measured by the 
sum of the series of trapezoids, or by the area ABC. This area 
is probably most conveniently measured by the trapezoids 
formed by the vertical lines representing solidities. The aver¬ 
age length of haul will be this area divided by the total solidity 
(represented in this case on p. 198 by the longest vertical line, 
2084). 

319. The construction of the “Mass Diagram” as a series 
of trapezoids involves the assumption that the center of gravity 
of a section of earthwork lies at its mid-section, which is only 
approximately correct since S for the first 50 ft. will seldom be 
exactly the same as S for the second 50 ft. of a section 100 ft. 
long. If the lines joining the ends of the vertical lines be made 
a curved line, the assumption becomes more closely accurate, 
and if the area be calculated by “Simpson’s Rule,” or by 
planimeter, results closely accurate will be reached. 

It will be further noticed that hill sections in the “ diagram ” 
represent haul forward on the profile, and valley sections haul 
backward. The mass diagram may therefore be used to indi¬ 
cate the methods by which the work shall be performed; 
whether excavation at any point shall be hauled forward or 
backward ; and, more particularly, to show the point where 
backward haul shall cease and forward haul begin, as indicated 
in the figure, p. 200, which shows a very simple case, the cuts 
and fills being evenly balanced, and no haul over 900 feet, with 
po rv^Pssitv for either borrowing 1 or wasting. 


PROfILE 


202 


Railroad Curves and Earthwork 


\ 





























Mass Diagram 


203 


320. In the figure, page 202, the excavation from Sta. 0 to 14 is 
very much in excess of embankment, and vice versa from Sta. 14 
to 30. The mass diagram indicates a haul of nearly 3000 ft. 
for a large mass of earthwork, measured by the ordinate AB. 
It will not be economical to haul the material 3000 ft. ; it is 
better to “ waste ” some of the material near Sta. 0, and to 
“ borrow ” some near Sta. 30, if this be possible, as is commonly 
the case. 

If we draw the line CD, the cut and fill between C and D will 
still be equal, and the volume of cut measured by CE can be 
wasted, and the equal volume of fill measured by DF can be 
borrowed to advantage. It can be seen that there is still a haul 
of nearly 2000 ft. (from A to D) on the large mass of earthwork 
measured by GH. It is probable that it will not pay to haul 
the mass GH, or any part of it, as far as AD. 

321. We must find the limit beyond which it is unprofitable 
to haul material rather than borrow and waste. 

Let c = cost of 1 cu. yd. excavation or embankment. 
h = cost of haul on 1 cu. yd. hauled 100 ft. 
n = length of haul in “ stations” of 100 ft. each. 

Then, when 1 cu. yd. of excavation is wasted, and 1 cu. yd. 
of embankment is borrowed, 

cost = 2 c 

When 1 cu. yd. of excavation is hauled into embankment, 

cost = c + nh 

The limit of profitable haul is reached when 

2 c = c + nh 



(209) 


or when 


Example. When excavation or embankment is 18 cents pel 
cu. yd., and haul is 1^ cents, 


18 

n — —- = 12 stations 


1.5 


When 


c = 16 and h — 2 
n — 8 stations 


then 


PROFILE 


204 


Railroad Curves and Earthwork. 































Mass Diagram. 


205 


322. In the former case (1200 ft., haul) we should draw in 
mass diagram (p. 204) the line KGL. Here KG is less than 
1200 ft. The line should not be lower than G, for in that case 
the haul would be nearly as great as KL, or more than 1200 ft. 

In the latter case (800 ft. haul) the line would be carried up 
to a point where NM = 800 ft. The masses between N and A, 
also C and O, can better be wasted than hauled, and the masses 
between M and G, also L and Z, can better be borrowed than 
hauled (always provided that there are suitable places at hand 
for borrowing and wasting). 

Next, produce NM to R. The number of yards borrowed 
will be the same whether taken at RZ or at MG -f LZ. That 
arrangement of work which gives the smallest “haul” (product 
of cu. yds. x distance hauled) is the best arrangement. The 
“haul ” in one case is measured by GLRYU, and in the other 
by MGU + UYR. If MGU is less than GLRU, then it is cheaper 
to borrow (a) RZ rather than (&) MG + LZ. 

In a similar way material NT and SO can be wasted more 
economically than NA and CO. 

The most economical position for the line MR is when MU = UR. 
For ST, when SV = VT. For any change from these positions of 
MR and ST will show an increase of area representing “ haul.” 

323. The case is often not as simple as that here given. 
Very often the material borrowed or wasted has to be hauled 
beyond the limit of “ free haul.” The limit beyond which it is 
unprofitable to haul will vary according to the length of haul 
on the borrowed or wasted material; the limit will, in general, 
be increased by the length of haul on the borrowed or wasted 
material. The haul on wasted or borrowed material, as NT, 
may be shown graphically by NTXW, where NW = TX shows 
the length of haul, and NTXW the “haul” (mass x distance). 

The mass diagram can be used also for finding the limit of 
“free haul” on the profile, and various applications will sug¬ 
gest themselves to those who become familiar with its use and 
the principles of its construction. Certainly one of its most 
important uses is in allowing “haul ” and “ borrow and waste” 
to be studied by a diagram giving a comprehensive view of the 
whole situation. There are few if any other available methods 
of accomplishing this result. 


206 


Railroad Carves and Earthwork. 


324. When material is first taken out in excavation, it 
generally occupies more space than was originally the case. 
When placed in embankment, it commonly shrinks somewhat 
and eventually occupies less space than originally. Wherever, 
from any cause, the material put into embankment will occupy j 
more space or less space than it did in excavation, the quantities 
in embankment should be corrected before figuring haul or con¬ 
structing a Mass Diagram, and an additional column should be 
shown in the Table, p. 197. 

325. Many engineers write their contracts and specifications 
without a clause allowing payment for “ haul ” or “overhaul.” 
Nevertheless it appears that it is the more common practice to 
insert a clause providing for payment for overhaul. A can¬ 
vass on this subject by the American Railway Engineering and 
Maintenance of Way Association in 1905 showed this practice 
to prevail in the proportion of 73 to 37. The free haul limit of 
500 ft. seemed to meet with greater favor than any other. 

Where an “ overhaul ” clause is inserted in a contract, the 
basis of payment has varied on different railroads. In one 
method, not recommended, the total haul is to be computed ; 
from this shall be deducted for free haul the total “yardage” 
multiplied by the length of the free haul limit. Under this sys¬ 
tem, with a 500 ft. free haul limit, there might be 10,000 cu. yds. 
of earth hauled (all of it) more than 500 ft., or an average of 
600 ft.; yet if there were another 10,000 cu. yds. hauled an 
average of 300 ft., there would be no payment whatever for 
overhaul; the average haul would be less than 500 ft. Un¬ 
less the specifications clearly show that this method is to be 
used, it is unfair as well as unsatisfactory to the contractor. 

What seems a logical and satisfactory provision is that 
lecominended by the American Railway Engineering and 
Maintenance of Way Association by a letter ballot vote of 134 
to 23 (announced in March, 1907). This is as follows : — 

“No payment will be made for hauling material when the 
length of haul does not exceed the limit of free haul, which 
shall be_feet. 

The limits of free haul shall be determined by fixing on the 
profile two points, one on each side of the neutral grade point, 
one in excavation and the other in embankment, such that the 
distance between thorn equals the specified free haul limi'. -nd 



Mass Diagram . 


20T 


the included quantities of excavation and embankment balance. 
All haul on material bevond this free haul limit will be esti¬ 
mated and paid for on the basis of the following method of 
computation, viz.: — 

“All material within this limit of free haul will be eliminated 
from further consideration. 

“The distance between the center of gravity of the remain¬ 
ing mass of excavation and center of gravity of the resulting- 
embankment, less the limit of free haul as above described, 
shall be the length of overhaul, and the compensation to be 
rendered therefor will be determined by multiplying the yard¬ 
age of the remaining mass as above described, by the length of 
the overhaul. Payment for the same will be by units .of one 
cubic yard hauled one hundred feet. 

“ When material is obtained from borrow-pits along the 
embankment, and runways are constructed, the haul shall be 
determined by the distance the team necessarily travels. The 
overhaul on material thus hauled shall be determined by mul¬ 
tiplying the yardage so hauled by one half the round distance 
made by the team less the free haul distance. The runways 
will be established by the engineer.” 

326. This statement as to the method of figuring overhaul 
is explained very simply by the Mass Diagram below. The 
length of AB is that of the free haul limit (500 ft. in figure). 
The free haul is shown in the area ACDBHA. The amount of 
overhaul to be paid for is shown in 3 parts, ACM, BDE, EGN. 


A * 

PROFILE 

i_i_i—i— 



B 1 

1 1 1 1 

1 1 1-1— 



MASS DIAGRAM 

















208 Railroad Curves and Earthwork. 


Table for Three-Level Sections. Base 14', Slope 1^ to 1. 



0 

.1 

.3 

.3 

.4 



L 

K 

L 

K 

L 

K 

L 

K 

L 

K 


0 

O. 

6-5 

2.6 

6.6 

5-3 

6.8 

8.0 

6.9 

10.8 

7.0 

0 

I 

28.7 

7-9 

3 i -9 

8.0 

35 - 1 

8.1 

38-4 

8-3 

41.7 

8.4 

1 

2 

63.0 

9-3 

66.7 

9.4 

7°-5 

9-5 

74-3 

9-7 

78.2 

9.8 

2 

3 

102.8 

10.6 

IO7.I 

10.8 

hi.4 

IO.9 

115.8 

1 I.I 

120.3 

II .2 

3 

4 

148.1 

12.0 

i 53 -o 

12.2 

157-9 

12.3 

162.8 

12.5 

167.9 

12.6 

4 

5 

199.1 

13-4 

204.5 

13.6 

209.9 

x 3-7 

215-4 

13-8 

221.0 

I4.O 

5 

6 

255-6 

14.8 

261.5 

15.0 

267.5 

i 5 -i 

273.6 

15.2 

279.7 

15-4 

6 

7 

3 x 7-6 

16.2 

324.1 

16.3 

330.7 

16.5 

337-3 

16.6 

344 -o 

16.8 

7 

8 

385-2 

17.6 

392.2 

17.7 

399-4 

17.9 

406.5 

18.0 

4 I 3-8 

18.1 

8 

9 

458.3 

I9.O 

466.0 

19.1 

473-6 

T 9-3 

481.4 

19.4 

489.1 

19-5 

9 

IO 

537 -o 

20.4 

545-2 

20.5 

553-4 

20.6 

561-7 

20.8 

570-1 

20.9 

10 

ii 

621.3 

21.8 

630.0 

21.9 

638.8 

22.0 

647.7 

22.2 

656.6 

22.3 

11 

12 

711.1 

23.1 

720 4 

23-3 

729.7 

23-4 

739-1 

23.6 

748.6 

23-7 

12 

13 

806.5 

24.5 

816.3 

24.7 

826.2 

24.8 

836.2 

25.0 

846.2 

25-1 

13 

14 

907.4 

25.9 

9 t 7-8 

26.1 

928.3 

26.2 

938.8 

26.3 

949-3 

26.5 

14 

15 

1013.9 

27-3 

1024.8 

27.5 

1035.9 

27.6 

1046.9 

27.7 

1058.0 

27.9 

15 

16 

1125-9 

28.7 

U 37-4 

28.8 

1149.0 

29.O 

1160.6 

29.I 

1172.3 

29-3 

l6 

17 

1243-5 

3 °-i 

1255-6 

30.2 

1267.7 

30-4 

1279.9 

30-5 

I292. I 

30.6 

17 

18 

1366.7 

3 i -5 

1379-3 

31.6 

1392.0 

31.8 

1404.7 

31-9 

1417-5 

32.0 

l8 

19 

1495.4 

32.9 

1508.5 

33 -o 

1521.8 

33- 1 

1535-1 

33-3 

1548.4 

33-4 

19 

20 

1629.6 

34-3 

1643.4 

34-4 

1657.1 

34-5 

1671.0 

34-7 

1684.9 

34-8 

20 


0 

.1 

.2 

.3 

.4 




.5 

.6 

.7 

.8 

.9 



L 

K 

L 

K 

L 

K 

L 

K 

L 

K 


O 

i3-7 

7-2 

16.6 

7-3 

19-5 

7-5 

22.5 

7.6 

25 6 

7-7 

0 

I 

45-i 

8.6 

48.6 

8.7 

52.1 

8.8 

55-7 

9.0 

59-3 

9.1 

1 

2 

82.2 

10.0 

86.2 

IO. I 

90.2 

10.2 

94.4 

IO.4 

98.5 

10.5 

2 

3 

124.8 

ii*3 

129.3 

ii -5 

1.34-0 

11.6 

138.6 

11.8 

143-4 

11.9 

3 

4 

172.9 

12.7 

178.0 

12.9 

183.2 

13.0 

188.4 

13-1 

193-7 

13-3 

4 

5 

226.6 

14.1 

232.3 

14-3 

238.0 

14.4 

243.8 

14 5 

249.7 

14.7 

5 

6 

285.9 

15-5 

292.1 

15.6 

298.4 

15-8 

304.7 

15-9 

311-1 

l6.I 

6 

7 

350.7 

16.9 

357-5 

17.0 

364-3 

17. 2 

371-2 

J7-3 

378.2 

17-5 

7 

8 

421.I 

18.3 

428.4 

18.4 

435-8 

18.6 

443-3 

18.7 

450.8 

18.8 

8 

9 

497.0 

19.7 

5°4-9 

19.8 

512.8 

20.0 

520.9 

20. I 

528.9 

20.2 

9 

10 

578-5 

21.1 

586.9 

21.2 

595-4 

21.3 

604.0 

21.5 

612.6 

21.6 

10 

11 

665-5 

22.5 

674-5 

22.6 

683.6 

22.7 

602.7 

22.9 

701.9 

23.0 

11 

12 

758.1 

23.8 

767.7 

24.O 

777-3 

24.1 

787.0 

24-3 

796.7 

24.4 

12 

13 

856.2 

25.2 

866.4 

25-4 

876.5 

25-5 

886.8 

25.6 

897.1 

25.8 

13 

14 

960.0 

26.6 

970.6 

26.8 

981.4 

26.9 

992.1 

27.O 

IOO3.O 

27.2 

14 

15 

1069.2 

28.0 

1080.4 

28.1 

IO9I.7 

28.3 

1103.1 

28.4 

IH4-5 

28.6 

15 

16 

1184.0 

29.4 

1195.8 

29-5 

1207.7 

29.7 

1219.6 

29.8 

I23I-5 

30.0 

16 

17 

1304-4 

30.8 

1316.7 

30-9 

1329-1 

3 1 -1 

1341.6 

31.2 

I354-I 

31-3 

17 

18 

1430.3 

32.2 

1443-2 

32-3 

1456.2 

32-5 

1469.2 

32.6 

1482.2 

3 2 -7 

18 

!9 

1561.8 

33-6 

1575-3 

33-7 

1588.8 

33-8 

1602.3 

34-o 

1616.0 

34.1 

19 

20 

1693.8 

35-o 

1712.9 

35-i 

1726.9 

35-2 

1741.0 

35-4 

1755-2 

35-5 

20 


.5 

.6 

.7 

.8 

.9 
















































































Tables 


209 


Table for Three-Level Sections. Base 20', Slope 1| to 1. 



0 


.1 


.2 


.3 


.4 




L 

K 

L 

K 

L 

K 

L 

I < 

L 

K 


o 

O. 

9-3 

3-7 

9-4 

7-5 

9-5 

11.4 

9-7 

15-3 

9.8 

0 

I 

39 - 8 

10.6 

44.1 

10.8 

48.4 

IO.9 

52.8 

I I.I 

57-3 

II .2 

1 

2 

85.2 

12.0 

90.0 

12.2 

94.9 

12.3 

99.9 

12.5 

104.9 

12.6 

2 

3 

136.1 

13-4 

Hi -5 

13.6 

147.0 

13-7 

152-5 

13.8 

158.0 

I4.O 

3 

4 

192.6 

14.8 

198.5 

15.0 

204.6 

I 5 - 1 

210.6 

15.2 

216.7 

15-4 

4 

5 

254.6 

16.2 

261.1 

16.3 

267.7 

16.5 

274-3 

16.6 

281.0 

16.8 

5 

6 

322.2 

17.6 

329-3 

17.7 

336-4 

17.9 

343-6 

18.0 

350-8 

l8.I 

6 

7 

395-4 

Iv^.O 

403.0 

19.1 

410.7 

x 9-3 

418.4 

19.4 

426.2 

19-5 

7 

8 

8 

47 V- 1 

20.4 

482.2 

20.5 

49°-5 

20.6 

498.8 

20.8 

5°7-i 

20.9 

q 

558.3 

21.8 

567- 1 

2I.9 

575-9 

22.0 

584-7 

22.2 

593-6 

22.3 

9 

IO 

648.1 

23.1 

657-4 

23.3 

666.8 

23-4 

676.2 

23.6 

685.6 

23-7 

10 

II 

743-5 

24-5 

753-4 

24.7 

763-3 

24.8 

773-2 

25.0 

783.2 

25 - 1 

11 

12 

844.4 

25-9 

8 5 4 - 8 

26.1 

865-3 

26.2 

875.8 

26.3 

886.4 

26.5 

12 

13 

950.9 

27.3 

961.9 

27-5 

972.9 

27.6 

984.0 

27.7 

995 - 1 

27.9 

13 

14 

10 >3.0 

28.7 

1074.5 

28.8 

1086.0 

29.0 

1097.7 

29.I 

1109.3 

29-3 

J 4 

15 

1180.6 

30.1 

1192.6 

30.2 

1204.7 

3°-4 

1216.9 

30-5 

1229.I 

30.6 

15 

l6 

1303.7 

31-5 

I 3 i6 -3 

3 1 -6 

1329.0 

3 i- 8 

I 34 I -7 

31-9 

1354-5 

32.0 

16 

17 

1432.4 

32.9 

1445.6 

33 ° 

1458.8 

33 -i 

1472.1 

33-3 

1485.4 

33 - 4 

34 - 8 
36.2 

17 

18 

18 

1566.7 

34-3 

1580.4 

34-4 

1594.2 

34-5 

1608.0 

34-7 

1621.9 

IQ 

1706.5 

35 - 6 

1720.8 

35 - 8 

I 735 - 1 

35-9 

1749-5 

36.1 

1764.0 

19 

20 

1851.9 

37 -o 

1866.7 

37-2 

1881.6 

| 37-3 

1896.5 

37-5 

1911.6 

37-6 

20 


0 

— 

.1 


.2 

> 

.3 

A 

L 




.5 

.6 

.7 

.8 

.9 



L 

K 

L 

K 

L 

I< 

L 

K 

L 

K 


0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

19 

20 

19.2 

61.8 

110.0 

163.7 
222 .() 

287.7 

358 -i 

434-o 

515-5 

602.5 

695.1 

793-3 

897.0 

1006.2 

1121.1 

1241.4 

1367.4 

1498.8 

1635.9 
I778-5 
1926.6 

10.0 

n -3 

12.7 

14.1 

I 5-5 

16.9 

18.3 

19.7 

21.1 

22.5 

23.8 

25.2 
26.6 
28.0 

29.4 

30.8 

32.2 
33-6 

35- o 

36- 3 

37- 7 

23.2 

66.4 

” 5 .i 

169.3 
229.1 
294-5 

365 4 
441.9 
524.0 

611.6 

704.7 

803.4 

907.7 
1017.5 
1132.9 

1253.8 

1380.3 

i5 i2 -3 

1649.9 
i793-o 
I94I-7 

10.1 

I”5 
12.9 
i4-3 

15.6 
17.0 

18.4 

19.8 

21.2 

22.6 
24.0 

25-4 

26.8 
28.1 

29.5 
3°-9 

3 2 - 3 

33- 7 
35-i 

36.5 

37-9 

27-3 

71.0 
120.2 

i75-i 
235-4 
3°i-4 

372.8 

449.9 

532.5 

620.6 

7I4-3 

813.6 
918.4 

1028.8 
H44-7 
1266.2 

1393-2 

1525-8 

1664.0 

1807.7 

1956.9 

10.2 

11.6 
13.0 
14.4 

15.8 

17.2 

18.6 
20.0 

21.3 

22.7 
24.1 

25-5 

26.9 

28.3 

29.7 

3 1 - ! 

32- 5 

33- 8 
35-2 
36.6 
38.0 

31-4 

75-7 

125-5 

180.9 

241.8 

308.3 

380.3 

457-9 

541.0 

629.7 
724.0 

823.8 
929.2 

1040.1 

1156.6 

1278.6 

1406.2 

1539-4 

1678.1 

1822.3 

1972.1 

10.4 

11.8 

13- i 

14- 5 

15- 9 
17-3 

18.7 

20.1 

21.5 

22.9 

24.3 

25.6 
27.0 

28.4 

29.8 

31.2 

32.6 

34- o 

35- 4 

36.8 
38.1 

35-6 

80.4 

130.8 

186.7 
248.2 
3*5-2 

387.8 
466.0 

549-7 

638.9 

733-7 

834.1 

940.0 

1051-5 

1168.5 

1291.1 

I4I9-3 

i553-o 

1692.2 
1837.1 
1987.4 

10.5 

11.9 

i3-3 

14.7 

16.1 
i7-5 

18.8 

20.2 

21.6 
23.0 
24.4 

25.8 

27.2 

28.6 
30.0 
3i-3 

32.7 

34- 1 

35- 5 

36.9 

38.3 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

19 

20 


.5 

.0 

.7 

.8 


1 -9 























































































210 


Railroad Curves and Earthwork 


Table of Prismoidal Corrections. 


C 0~ C 1 

1 

2 

3 

4 

5 

G 

7 

8 

9 

c 0 — c l 

A-A 

O.I 

•03 

.06 

.09 

.12 

•15 

.19 

.22 

.25 

.28 

A-A 

0.1 

.2 

.06 

.12 

.19 

.25 

• 3 i 

•37 

•43 

•49 

•56 

.2 

•3 

.09 

.19 

.28 

•37 

.46 

•56 

•65 

•74 

•83 

•3 

•4 

.12 

.25 

•37 

•49 

.62 

•74 

.86 

•99 

1.11 

•4 

•5 

•15 

•31 

.46 

.62 

•77 

•93 

1.08 

1.23 

i -39 

•5 

.6 

.19 

•37 

•56 

•74 

•93 

I.II 

1.30 

1.48 

1.67 

.6 

•7 

.22 

•43 

•65 

.86 

i.08 

1.30 

I * 5 I 

i -73 

1.94 

•7 

.8 

•25 

•49 

•74 

•99 

1.23 

1.48 

i -73 

1.98 

2.22 

.8 

•9 

.28 

•56 

.83 

1.11 

i -39 

1.67 

1.94 

2.22 

2.50 

•9 

1.0 

•31 

.62 

•93 

1.23 

i -54 

1.85 

2.16 

2.47 

2.78 

1.0 

.1 

•34 

.68 

1.02 

1.36 

1.70 

2.04 

2.38 

2.72 

3.06 

.1 

.2 

•37 

•74 

I.II 

1.48 

1.85 

2.22 

2-59 

2.96 

3-33 

.2 

•3 

.40 

,80 

1.20 

1.60 

2.01 

2.41 

2.81 

3-21 

3.61 

•3 

4 

•43 

.86 

1.30 

i -73 

2.16 

2.59 

3.02 

3-46 

3-89 

•4 

•5 

.46 

•93 

I -39 

1.85 

2.31 

2.78 

3-24 

3 - 7 ° 

4.17 

•5 

.6 

.49 

•99 

i-48 

1.98 

2.47 

2.96 

3-46 

3-95 

4.44 

.6 

•7 

•52 

1.05 

I -57 

2.10 

2.62 

3-15 

3-67 

4.20 

4.72 

•7 

.8 

•56 

I.II 

1.67 

2.22 

2.78 

3-33 

3-89 

4-44 

5.00 

.8 

•9 

•59 

1.17 

1.76 

2-35 

2.93 

3-52 

4.10 

4.69 

5.28 

•9 

2.0 

.62 

1.23 

1.85 

2-47 

3-09 

3 - 7 ° 

4 - 3 2 

4-94 

5-56 

2.0 

.1 

•65 

1.30 

i -94 

2.59 

3-24 

3-89 

4-54 

5 -i 9 

5-83 

.1 

.2 

.68 

1.36 

2.04 

2.72 

3-40 

4.07 

4-75 

5-43 

6.ii 

.2 

•3 

•71 

I .42 

2.13 

2.84 

3-55 

4.26 

4 97 

5-68 

6-39 

•3 

•4 

•74 

1.48 

2.22 

2.96 

3-70 

4-44 

5 -i 9 

5-93 

6.67 

4 

•5 

•77 

i -54 

2.31 

3°9 

3.86 

4-^3 

5 40 

6.17 

6.94 

•5 

.6 

.80 

1.60 

2.4I 

3.21 

4.01 

4.81 

5.62 

6.42 

7.22 

.6 

•7 

•83 

1.67 

2.50 

3-33 

4 -i 7 

5.00 

5-83 

6.67 

7 - 5 ° 

•7 

.8 

.86 

*•73 

2 - 5 J 

3-46 

4-32 

5-19 

6.05 

6.91 

7.78 

.8 

•9 

.90 

T -79 

2.69 

3-58 

4.48 

5-37 

6.27 

7.16 

8.06 

•9 

30 

•93 

1.85 

2.78 

3-70 

4-63 

5-56 

6.48 

7.41 

8-33 

30 

.1 

.96 

1.91 

2.87 

3-83 

4.78 

5-74 

6.70 

7-65 

8.61 

.1 

.2 

•99 

1.98 

2.96 

3-95 

4-94 

5-93 

6.91 

7.90 

8.89 

.2 

•3 

1.02 

2.04 

3.06 

4.07 

5-09 

6. ii 

7-»3 

8.15 

9.17 

•3 

•4 

1.05 

2 . IO 

3-15 

4.20 

5-25 

6.30 

7-35 

8.40 

9-44 

■4 

•5 

1.08 

2.16 

3-24 

4-32 

5-40 

6.48 

7-56 

8.64 

9.72 

•5 

.6 

I.II 

2.22 

3-33 

4-44 

5-56 

6.67 

7.78 

8.89 

10.00 

.6 

•7 

1.14 

2.28 

3-43 

4-57 

5 - 7 1 

6.85 

7-99 

9.14 

10.28 

•7 

.8 

I * I 7 

2-35 

3 * 5 2 

4.69 

5.86 

7.04 

8.21 

9-38 

10.56 

.8 

•9 

1.20 

2.41 

3.61 

4.81 

6.02 

7.22 

8-43 

963 

10.83 

•9 

4.0 

1.23 

2.47 

3-70 

4-94 

6.17 

7 - 4 i 

8.64 

9.88 

II.II 

4.0 

.1 

I.27 

2-53 

3.80 

5.06 

6-33 

7-59 

8.86 

10.12 

n -39 

.1 

.2 

1.30 

2-59 

3-89 

5 -i 9 

6.48 

7.78 

9.07 

IO - 37 

11.67 

.2 

•3 

i -33 

2.65 

3-98 

5 - 3 i 

6.64 

7.96 

9.29 

10.62 

11.94 

•3 

•4 

1.36 

2.72 

4.07 

5-43 

6.79 

8.15 

9 -S 1 

10.86 

12.22 

•4 

•5 

i -39 

2.78 

4.17 

5-56 

6.94 

8-33 

9.72 

II.II 

12.50 

•5 

.6 

1.42 

2.84 

4.26 

5-68 

7.10 

8.52 

9.94 

11.36 

12.78 

.6 

•7 

I -45 

2.90 

4-35 

5.80 

7-25 

8.70 

10.15 

11.60 

13.06 

•7 

.8 

1.48 

2.96 

4.44 

5-93 

7.41 

8.89 

10 -37 i 

11.85 

13-33 

.8 

•9 

i- 5 r 

3.02 

4-54 

6.05 

7-56 

9.07 

10.59 

12.10 

13.61 

•9 

5 -° 

i -54 

3-09 

4-63 

6.17 

7.72 

9.26 

10.80 

12.35 

13.89 

5 -o 

<4 

© 

1 

1 

2 

3 

4 

5 

6 

7 

8 

9 

1 

O 

















































Tables 


211 


Table of Prismoidal Corrections. 


© 

1 

a 

1 

2 

3 

4 

5 

6 

7 

8 

9 

C 0 —Cj 

A-A 

5 -i 

i -57 

3 -i 5 

4.72 

6.30 

7.87 

9.44 

11.02 

12.59 

14.17 

A-A 

5-1 

.2 

i.6o 

3.21 

4.81 

6.42 

8.02 

9-63 

n.23 

12.84 

14.44 

.2 

•3 

1.64 

3.27 

4.91 

6-54 

8.18 

9.81 

n -45 

13.09 

14.72 

•3 

•4 

1.67 

3-33 

5.00 

6.67 

8.33 

10.00 

11.67 

13-33 

15.0° 

•4 

•5 

1.70 

3-40 

5-09 

6.79 

8.49 

10.19 

11.88 

13.58 

15.28 

•5 

.6 

i -73 

3-46 

5 - 1 9 

6.91 

8.64 

10.37 

12.10 

13-83 

15-56 

.6 

•7 

1.76 

3-52 

5.28 

7.04 

8.80 

10.56 

12.31 

14.07 

1583 

•7 

.8 

1.79 

3-58 

5-37 

7.16 

8.95 

10.74 

12.53 

14-32 

16.11 

.8 

•9 

1.82 

3-64 

5-46 

7.28 

9.IO 

io -93 

12.75 

14-57 

16.39 

•9 

6.0 

1.85 

3 - 7 ° 

5-56 

7.41 

9.26 

II.II 

12.96 

14.81 

16.67 

6.0 

.1 

1.88 

3-77 

5-65 

7-53 

9.41 

11.30 

13.18 

15.06 

16.94 

.1 

.2 

1.91 

3-83 

5-74 

7-65 

9-57 

11.48 

13.40 

i 5 - 3 i 

17.22 

.2 

•3 

1.94 

3-89 

5-83 

7.78 

9.72 

11.67 

13.61 

15-56 

17-50 

■3 

•4 

1.98 

3-95 

5-93 

7.90 

9.88 

11.85 

13-83 

15.80 

17.78 

■4 

•5 

2.01 

4.01 

6.02 

8.02 

10.03 

12.04 

14.04 

16.05 

18.06 

•5 

.6 

2.O4 

4-°7 

6.ii 

8.15 

IO.I9 

12.22 

14.26 

16.30 

18.33 

.6 

•7 

2.O7 

4.14 

6.20 

8.27 

10.34 

I2.4I 

14.48 

16.54 

18.61 

•7 

.8 

2 . IO 

4.20 

6.30 

8.40 

10.49 

12.59 

14.69 

16.79 

18.89 

.8 

•9 

2.13 

4.26 

6-39 

8.52 

10.65 

12.78 

14.91 

17.04 

19.17 

■9 

7.0 

2.16 

4-32 

6.48 

8.64 

10.80 

12.96 

15.12 

17.28 

19.44 

7.0 

.i 

2.I9 

4-38 

6-57 

8.77 

10.96 

13-15 

15-34 

17-53 

19.72 

.1 

.2 

2.22 

4.44 

6.67 

8.89 

II.II 

13-33 

15-56 

17.78 

20.00 

.2 

3 

2.25 

4-51 

6.76 

9.01 

11.27 

13-52 

15-77 

18.02 

20.28 

•3 

•4 

2.28 

4-57 

6.85 

9.14 

11.42 

13-70 

15-99 

18.27 

20.56 

■4 

•5 

2.31 

4-63 

6.94 

9.26 

n -57 

13.89 

16.20 

18.52 

20.83 

•5 

.6 

2-35 

4.69 

7.04 

9.38 

H -73 

14.07 

16.42 

18.77 

21 .II 

.6 

•7 

2.38 

4-75 

7 -i 3 

9 - 5 1 

11.88 

14.26 

16.64 

19.01 

21.39 

•7 

.8 

2.41 

4.81 

7.22 

9-63 

12.04 

14.44 

16.85 

19.26 

21.67 

.8 

•9 

2.44 

4.88 

7 - 3 i 

9-75 

12.19 

14.63 

17.07 

i 9 - 5 i 

21.94 

•9 

8.o 

2.47 

4.94 

7 - 4 1 

9.88 

12.35 

14.81 

17.28 

19-75 

22.22 

8.0 

.1 

2.50 

5.00 

7-50 

10.00 

12.50 

15.00 

17-50 

20.00 

22.50 

.1 

.2 

2-53 

5.06 

7-59 

10.12 

12.65 

15-19 

17.72 

20.25 

22.78 

.2 

•3 

2.56 

5.12 

7.69 

10.25 

12.81 

15-37 

17-93 

20.49 

23.06 

•3 

•4 

2-59 

S-i 9 

7.78 

10.37 

12.96 

15-56 

18.15 

20.74 

23-33 

•4 

•5 

2.62 

5-25 

7.87 

10.49 

i 3 - 12 

15-74 

18.36 

2O.99 

23.61 

•5 

.6 

2.63 

5 - 3 i 

7.96 

10.62 

13-27 

15-93 

18.58 

21.23 

23.89 

.6 

• 7 

2.69 

5-37 

8.06 

10.74 

13-43 

16.11 

18.80 

21.48 

24.17 

•7 

.8 

2.72 

5-43 

8.15 

10.86 

I 3-58 

16.30 

19.01 

21-73 

24.44 

.8 

•9 

2-75 

5-49 

8.24 

10.99 

13-73 

16.48 

19.23 

21.97 

24.72 

•9 

9 '° 

2.78 

5-56 

8-33 

II.II 

13.89 

16.67 

19.44 

22.22 

25.00 

9.0 

.1 

2.81 

5.62 

8-43 

n.23 

14.04 

16.85 

19.66 

22.47 

25.28 

.1 

.2 

2.84 

5-68 

8.52 

n.36 

14.20 

17.04 

19.88 

22.72 

25-56 

.2 

•3 

2.87 

5-74 

8.61 

11.48 

14-35 

17.22 

20.09 

22.96 

25-83 

•3 

•4 

2.90 

5.80 

8.70 

11.60 

M- 5 I 

17.41 

20.31 

23.21 

26.11 

•4 

•5 

2-93 

5-86 

8.80 

n -73 

14.66 

17-59 

20.52 

23.46 

26.39 

•5 

.6 

2.96 

5-93 

8.89 

11.85 

M 

CO 

H 

17.78 

20.74 

23.70 

26.67 

.6 

•7 

2.99 

5-99 

8.98 

11.98 

14.97 

17.96 

20.96 

23-95 

26.94 

•7 

.8 

.8 

3.02 

6.05 

9.07 

12 . IO 

15.12 

18.15 

21.17 

24.20 

27.22 

•9 

3.06 

6.11 

Q.17 

12.22 

15.28 

18.33 

21.39 

24.44 

27.50 

■9 

10.0 

3-09 

6.17 

9.26 

12.35 

15-43 

18.52 

21.60 

24.69 

27.78 

10.0 

o 

1 

£ 

1 

2 

3 

4 

5 

6 

7 

8 

9 

c 0 —c t 











































212 


Railroad Curves and Earthwork 


ALLEN’S TABLES (Copyright, 1893, by C. F. Allen). 

Triangular Prisms. S in cu. yds. for 50 ft. in length. 


WIDTH 

1 

2 

3 

4 

5 

6 

7 

8 

9 

WIDTH 

Height 

0.1 

.09 

.19 

.28 

•37 

.46 

• 56 

•65 

•74 

.83 

Height 

0.1 

.2 

.19 

•37 

•56 

•74 

•93 

I.II 

1.30 

1.48 

1.67 

.2 

•3 

.28 

•56 

.83 

I.II 

i -39 

1.67 

1.94 

2.22 

2.50 

•3 

•4 

•37 

•74 

I.II 

1.48 

1.85 

2.22 

2.59 

2.96 

3-33 

•4 

•5 

.46 

•93 

i -39 

1.85 

2.31 

2.78 

3-24 

3 - 7 ° 

4.17 

■5 

.6 

•56 

I.II 

1.67 

2.22 

2.78 

3-33 

3- 8 9 

4.44 

5.00 

.6 

•7 

•65 

1.30 

1.94 

2-59 

3-24 

3-89 

4-54 

5 -i 9 

5 - 8 3 

•7 

.8 

•74 

1.48 

2.22 

2.96 

3-70 

4-44 

5 -i 9 

5-93 

6.67 

.8 

•9 

.83 

1.67 

2.50 

3-33 

4 -i 7 

5.00 

5.83 

6.67 

7 - 5 o 

•9 

1.0 

•93 

1.85 

2.78 

3-70 

4-63 

5-56 

6.48 

7.41 

8-33 

1.0 

.i 

1.02 

2.04 

3.06 

4.07 

5-09 

6. ii 

7 -i 3 

8.15 

9.17 

.1 

.2 

I.1I 

2.22 - 

3-33 

4.44 

5-56 

6.67 

7.78 

8.89 

10.00 

.2 

■3 

1.20 

2.4I 

3.61 

4.81 

6.02 

7.22 

8-43 

9-63 

10.83 

•3 

•4 

1.30 

2-59 

3-89 

5 -i 9 

6.48 

7.78 

9-°7 

IO -37 

11.67 

•4 

•5 

1 *39 

2.78 

4.17 

5-56 

6.94 

8-33 

9.72 

II.II 

12.50 

•5 

.6 

1.48 

2.96 

4.44 

5-93 

7.41 

8.89 

IO -37 

11.85 

1 3-33 

.6 

•7 

I> 57 

3 -i 5 

4.72 

6.30 

7.87 

9-44 

11.02 

12.59 

14.17 

•7 

.8 

1.67 

3-33 

5.00 

6.67 

8-33 

10.00 

n.67 

13-33 

15.00 

.8 

•9 

1.76 

3-52 

5.28 

7.04 

8.80 

10.56 

12.31 

14.07 

15-83 

•9 

2.0 

1.85 

3 - 7 » 

5-56 

7 - 4 i 

9.26 

II.II 

12.96 

14.81 

16.67 

2.0 

.1 

1.94 

3-89 

5.83 

7.78 

9.72 

11.67 

13.61 

15-56 

1 7 - 5 ° 

.1 

.2 

2.04 

4.07 

6.ii 

8.15 

IO.I9 

12.22 

14.26 

16.30 

18.33 

.2 

•3 

2.13 

4.26 

6-39 

8.52 

10.65 

12.78 

14.91 

17.04 

19.17 

■3 

4 

2.22 

4.44 

6.67 

8.89 

II.II 

* 3-33 

15-56 

17.78 

20.00 

•4 

•5 

2.31 

4-63 

6.94 

9.26 

11 -57 

13.89 

16.20 

18.52 

20.83 

•5 

.6 

2.41 

4.81 

7.22 

9-63 

12.04 

14.44 

16.85 

19.26 

21.67 

.6 

•7 

2.50 

5.00 

7-50 

10.00 

12.50 

15.00 

17-50 

20.00 

22.50 

•7 

.8 

2-59 

5 -i 9 

7.78 

IO -37 

12.96 

15-56 

18.15 

20.74 

23-33 

.8 

•9 

2.69 

5-37 

8.06 

10.74 

13-43 

16.11 

18.80 

21.48 

24.17 

•9 

3-0 

2.78 

5-56 

8-33 

II.II 

13.89 

16.67 

19.44 

22.22 

25.00 

3 0 

.1 

2.87 

5-74 

8.61 

11.48 

14-35 

17.22 

20.09 

22.96 

25-83 

.1 

.2 

2.96 

5-93 

8.89 

11.85 

14.81 

17.78 

20.74 

23.70 

26.67 

.2 

-3 

3.06 

6.ii 

9.17 

12.22 

15.28 

j 8-33 

21.39 

24.44 

27.50 

•3 

•4 

3 -i 5 

6.30 

9.44 

12.59 

15-74 

18.89 

22.04 

25.19 

28.33 

■4 

•5 

3-24 

6.48 

9.72 

12.96 

16.20 

19.44 

22.69 

25-93 

29.17 

•5 

.6 

3.33 

6.67 

10.00 

1333 

16.67 

20.00 

23-33 

26.67 

30.00 

.6 

•7 

3-43 

6.85 

10.28 

13-7° 

17-13 

20.56 

23.98 

27.41 

30.83 

•7 

.8 

3-52 

7.04 

10.56 

14.07 

J 7-59 

21.11 

24.63 

28.15 

31-67 

.8 

•9 

3.61 

7.22 

10.83 

14.44 

18.06 

21.67 

25.28 

28.89 

32.50 

•9 

4.0 

3 - 7 ° 

7.41 

II.II 

14.81 

18.52 

22.22 

25-93 

29.63 

33-33 

4.0 

.1 

3.80 

7-59 

n -39 

15-19 

18.98 

22.78 

26.57 

30.37 

34-17 

.1 

.2 

3-89 

7.78 

11.67 

15-56 

19.44 

23-33 

27.22 

31.11 

35 00 

.2 

•3 

3-98 

7.96 

n.94 

15-93 

I 9 - 9 I 

23.89 

27.87 

31-85 

35-83 

•3 

•4 

4.07 

8.15 

12.22 

16.30. 

20.37 

24.44 

28.52 

32.59 

36.67 

•4 

•5 

4.17 

8-33 

12.50 

16.67 

20.83 

25.00 

29.17 

33-33 

37-50 

•5 

.6 

4.26 

8.52 

12.78 

I 7-°4 

21.30 

25-56 

29.81 

34-07 

38-33 

.6 

■7 

4-33 

8.70 

13.06 

17.41 

21.76 

26.11 

30.46 

34.81 

39-*7 

•7 

.8 

4.44 

8.89 

1 3-33 

17.78 

22.22 

26.67 

31.11 

35-56 

40.00 

.8 

•9 

4-54 

9-°7 

13.61 

18.15 

22.69 

27.22 

3!-76 

36.30 

40.83 

•9 

5-0 

4-63 

9.26 

13.89 

18.52 

23-15 

27.78 

32.41 

37-°4 

41.67 

5-0 


1 

2 

0 

3 

4 

5 

6 

7 

8 

9 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